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'A ball is thrown at an angle of 45 degrees to hit an object 40.82m away. What should be the initial velocity?'

Any help would be appreciated

Any help would be appreciated

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#2

Resolve the velocity into horizontal and vertical components. Then Suvat and solve simultaneously

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#3

Assume g= 9.8N/kg,

The horizontal component of the speed won't change (because there's no force on the horizontal axis, assuming no air resistance).

The vertical component of its speed will decrease and eventually turn negative due to the force of gravitational attraction between the ball and the earth.

At 45 degrees, the horizontal component and vertical component of its speed will be the same immediately after it is thrown (if you don't believe me, tan(45deg) = 1)

If you let x be the initial vertical component of speed, the time it travelled until its speed is 0 in the air can be calculated using x/9.8. The horizontal distance it travelled in this time is just the (constant horizontal component of speed) * time: x * x / 9.8 = x^2 / 9.8.

To figure out how high it is in the air, v^2 = u^2 + 2as, v is just 0, so you have s=x^2/19.6, x^2/19.6 = 0.05102x^2 .

Now to see how long until it hits the ground s = ut + 0.5at^2, u is just 0 because it has 0 velocity at its highest point in the air s= 0.5at^2, t= sqrt(s/0.5a),

t = sqrt(0.05102x^2 / 4.9)

= 0.102x (3sf).

So now let's see how far it flies horizontally during it's descent, same formula as above. 0.102x * x = 0.102x^2

So now the total horizontal distance flown is 0.102x^2 + x^2/9.8 = 40.82 (given in question)

x^2(0.102+1/9.8) = 40.82

x= sqrt(40.82/(0.102+1/9.8))

=14.144 (3dp)

But that's not its initial speed, just it's horizontal/vertical components of its initial speed. To find its initial speed, it's just pythagoras

sqrt(14.144^2 + 14.144^2) = 20.00m/s

If you get an exact answer like that, you probably got it right. At least I hope I did.

There's probably an easier way to do it than I did, but I'm only GCSE, so I haven't learned the "cleanest" way to do it.

The horizontal component of the speed won't change (because there's no force on the horizontal axis, assuming no air resistance).

The vertical component of its speed will decrease and eventually turn negative due to the force of gravitational attraction between the ball and the earth.

At 45 degrees, the horizontal component and vertical component of its speed will be the same immediately after it is thrown (if you don't believe me, tan(45deg) = 1)

If you let x be the initial vertical component of speed, the time it travelled until its speed is 0 in the air can be calculated using x/9.8. The horizontal distance it travelled in this time is just the (constant horizontal component of speed) * time: x * x / 9.8 = x^2 / 9.8.

To figure out how high it is in the air, v^2 = u^2 + 2as, v is just 0, so you have s=x^2/19.6, x^2/19.6 = 0.05102x^2 .

Now to see how long until it hits the ground s = ut + 0.5at^2, u is just 0 because it has 0 velocity at its highest point in the air s= 0.5at^2, t= sqrt(s/0.5a),

t = sqrt(0.05102x^2 / 4.9)

= 0.102x (3sf).

So now let's see how far it flies horizontally during it's descent, same formula as above. 0.102x * x = 0.102x^2

So now the total horizontal distance flown is 0.102x^2 + x^2/9.8 = 40.82 (given in question)

x^2(0.102+1/9.8) = 40.82

x= sqrt(40.82/(0.102+1/9.8))

=14.144 (3dp)

But that's not its initial speed, just it's horizontal/vertical components of its initial speed. To find its initial speed, it's just pythagoras

sqrt(14.144^2 + 14.144^2) = 20.00m/s

If you get an exact answer like that, you probably got it right. At least I hope I did.

There's probably an easier way to do it than I did, but I'm only GCSE, so I haven't learned the "cleanest" way to do it.

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#4

(Original post by

Assume g= 9.8N/kg,

The horizontal component of the speed won't change (because there's no force on the horizontal axis, assuming no air resistance).

The vertical component of its speed will decrease and eventually turn negative due to the force of gravitational attraction between the ball and the earth.

At 45 degrees, the horizontal component and vertical component of its speed will be the same immediately after it is thrown (if you don't believe me, tan(45deg) = 1)

If you let x be the initial vertical component of speed, the time it travelled until its speed is 0 in the air can be calculated using x/9.8. The horizontal distance it travelled in this time is just the (constant horizontal component of speed) * time: x * x / 9.8 = x^2 / 9.8.

To figure out how high it is in the air, v^2 = u^2 + 2as, v is just 0, so you have s=x^2/19.6, x^2/19.6 = 0.05102x^2 .

Now to see how long until it hits the ground s = ut + 0.5at^2, u is just 0 because it has 0 velocity at its highest point in the air s= 0.5at^2, t= sqrt(s/0.5a),

t = sqrt(0.05102x^2 / 4.9)

= 0.102x (3sf).

So now let's see how far it flies horizontally during it's descent, same formula as above. 0.102x * x = 0.102x^2

So now the total horizontal distance flown is 0.102x^2 + x^2/9.8 = 40.82 (given in question)

x^2(0.102+1/9.8) = 40.82

x= sqrt(40.82/(0.102+1/9.8))

=14.144 (3dp)

But that's not its initial speed, just it's horizontal/vertical components of its initial speed. To find its initial speed, it's just pythagoras

sqrt(14.144^2 + 14.144^2) = 20.00m/s

If you get an exact answer like that, you probably got it right. At least I hope I did.

There's probably an easier way to do it than I did, but I'm only GCSE, so I haven't learned the "cleanest" way to do it.

**Dench,x,Kid**)Assume g= 9.8N/kg,

The horizontal component of the speed won't change (because there's no force on the horizontal axis, assuming no air resistance).

The vertical component of its speed will decrease and eventually turn negative due to the force of gravitational attraction between the ball and the earth.

At 45 degrees, the horizontal component and vertical component of its speed will be the same immediately after it is thrown (if you don't believe me, tan(45deg) = 1)

If you let x be the initial vertical component of speed, the time it travelled until its speed is 0 in the air can be calculated using x/9.8. The horizontal distance it travelled in this time is just the (constant horizontal component of speed) * time: x * x / 9.8 = x^2 / 9.8.

To figure out how high it is in the air, v^2 = u^2 + 2as, v is just 0, so you have s=x^2/19.6, x^2/19.6 = 0.05102x^2 .

Now to see how long until it hits the ground s = ut + 0.5at^2, u is just 0 because it has 0 velocity at its highest point in the air s= 0.5at^2, t= sqrt(s/0.5a),

t = sqrt(0.05102x^2 / 4.9)

= 0.102x (3sf).

So now let's see how far it flies horizontally during it's descent, same formula as above. 0.102x * x = 0.102x^2

So now the total horizontal distance flown is 0.102x^2 + x^2/9.8 = 40.82 (given in question)

x^2(0.102+1/9.8) = 40.82

x= sqrt(40.82/(0.102+1/9.8))

=14.144 (3dp)

But that's not its initial speed, just it's horizontal/vertical components of its initial speed. To find its initial speed, it's just pythagoras

sqrt(14.144^2 + 14.144^2) = 20.00m/s

If you get an exact answer like that, you probably got it right. At least I hope I did.

There's probably an easier way to do it than I did, but I'm only GCSE, so I haven't learned the "cleanest" way to do it.

Answering requests should not be an exercise in telling the OP and everyone else that you know how to get the answer.

Thanks.

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#5

you can express y in terms of x....

y = xtan45° - 4.9x

we know y = 0, x = 40.82, so you can solve for V

y = xtan45° - 4.9x

^{2}/{v^{2}cos^{2}45°}we know y = 0, x = 40.82, so you can solve for V

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#6

(Original post by

'A ball is thrown at an angle of 45 degrees to hit an object 40.82m away. What should be the initial velocity?'

Any help would be appreciated

**Jayc3**)'A ball is thrown at an angle of 45 degrees to hit an object 40.82m away. What should be the initial velocity?'

Any help would be appreciated

On landing, y = 0, t > 0: 0 = [email protected] - 0.5gt

Sub in our expression for t, re-arrange and solve for v.

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