Year 12s: what will be the first way you research your future options? >>

Projectile Motion Q

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

Reply 1

Notnek

Can you please post your working?

Reply 2

TSR George

Original post by Notnek

Can you please post your working?

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

Reply 3

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).