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Projectile Motion Q watch

1. Hi guys

Can anyone help me with this question? I can't seem to get the answer in the back of the book.

It asks us to calculate the maximum height which I got correct - that was 23m approximately. It then asked us to calculate the horizontal displacement when the object lands on the ground. I got 45.88m.

We were only given the information that the golf ball has a velocity of 30ms^'1 at an angle of 45 degrees to the horizontal ground.

Any help would be greatly appreciated.
Thanks
2. (Original post by Blake Jones)
Hi guys

Can anyone help me with this question? I can't seem to get the answer in the back of the book.

It asks us to calculate the maximum height which I got correct - that was 23m approximately. It then asked us to calculate the horizontal displacement when the object lands on the ground. I got 45.88m.

We were only given the information that the golf ball has a velocity of 30ms^'1 at an angle of 45 degrees to the horizontal ground.

Any help would be greatly appreciated.
Thanks
3. (Original post by Notnek)
Sure, I got the velocity to be 21.21ms^-1 for both components.

I then used v^2 = u^2 +2as where v=0 a= -9.81 and u = 21.21
0 = 450 - 19.62s
s = 450 / 19.62 which gave me 22.94m

Then I used s=ut+0.5at^2 to get the time for the length of the whole projectile (so like how long it took to come back down to the ground)

22.94 = 21.21t because a=0 at the maximum height
therefore t=22.94/21.21 = 1.082 to get to max height
I then doubled this for the time taken for the full projectile which gave me 2.1631

Using s=ut (because a=0 horizontally) I then got s = 21.21 * 2.1631 = 45.88m
4. (Original post by Blake Jones)
22.94 = 21.21t because a=0 at the maximum height
This is incorrect. The acceleration due to gravity for the vertical motion is a constant : g = 9.8. It won't change throughout the vertical motion. The velocity does change and it is this that is equal to 0 at the apex.

I recommend using s = ut + 1/2 at^2 with s = 0. This will give you the total time of flight (since when the particle hits the ground its vertical displacement will be 0).
5. (Original post by Notnek)
This is incorrect. The acceleration due to gravity for the vertical motion is a constant : g = 9.8. It won't change throughout the vertical motion. The velocity does change and it is this that is equal to 0 at the apex.

I recommend using s = ut + 1/2 at^2 with s = 0. This will give you the total time of flight (since when the particle hits the ground its vertical displacement will be 0).
Oh! I see! Thank you!

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