The Student Room Group

projectiles question

A ball is kicked so that it acquires an initial velocity of 5ms^(-1) at an angle of 45 degrees to the horizontal.

What height will it reach?
How long will it take before the ball hits the ground?
How far will it travel horizontally before it hits the ground?


i) for the height I used the vertical component and got s=0.63
ii) I know I'd have to consider the horizontal component but would that mean s=0.63 x2 assuming the flight path is symmetrical, and then use s=ut to find the time ?
iii) Once I calculate the time from ii) I can sub it into s=ut +1/2at^2 ?
the height 0.63m seems small but is actually correct ?
(edited 8 years ago)
for part ii) you can work with the vertical.distance... it must equal zero.

0 = 5sin45ºt + 1/2*(-9.8)*t2

this is quadratic...

iii) use the answer from ii) in s = ut + 1/2*0*t2

the acceleration is zero horizontally...
Original post by the bear
the height 0.63m seems small but is actually correct ?

We don't have access to a mark scheme unfortunately
Original post by the bear
for part ii) you can work with the vertical.distance... it must equal zero.

0 = 5sin45ºt + 1/2*(-9.8)*t2

this is quadratic...

iii) use the answer from ii) in s = ut + 1/2*0*t2

the acceleration is zero horizontally...


for part ii) could I use the total distance as 0.63 x 2 ? would that still work if I used the horizontal component ?
Original post by T-GiuR
for part ii) could I use the total distance as 0.63 x 2 ? would that still work if I used the horizontal component ?


the total distance moved vertically is 0.63*2 metres, but that is not S in the formulas... S is the displacement... so at the end of the flight the vertical displacement is zero.
Original post by the bear
the total distance moved vertically is 0.63*2 metres, but that is not S in the formulas... S is the displacement... so at the end of the flight the vertical displacement is zero.

I thought it would of been the total distance horizontally ?
If the max height is achieved at 0.63m, which is symmetrically in the middle ?
Original post by T-GiuR
I thought it would of been the total distance horizontally ?
If the max height is achieved at 0.63m, which is symmetrically in the middle ?


you cannot find the total time using the horizontal velocity... you use the vertical velocity instead. then find the horizontal distance using the total time and the horizontal component of velocity.
Original post by the bear
you cannot find the total time using the horizontal velocity... you use the vertical velocity instead. then find the horizontal distance using the total time and the horizontal component of velocity.

Understood. Thanks :smile:

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