Projectile motion Q (1 Viewer)

[passion89]

I just have a quick projectile motion question which I need to check the answers for:

A golfer strikes a ball on the ground on a level golf course. The ball hits the ground 180 metres north from where it was struck, 5.6 seconds later. Find:
a) The maximum height the ball reached
b) The initial velocity (direction, angle and speed) of the ball as it left the club

 

[Riviet]

a)Since time of flight is 5.6s, time to reach max height is 2.8s and when max height is reached, v=0.

v=u+at
0=u-9.8x2.8
u_y=27.44m/s (vertical component)

s=ut + 1/2.at²
=(27.44)(2.8) - 1/2.(9.8)(2.8)²
=38.416
=38m
.'. max height reached is 38m above ground.

b) Using distance (d) = vt
u_x=d/t
=180/5.6
=32m/s (horizontal component)

Use Pythagora's Theorem with the vertical and horizontal components of the initial velocity u (in red) to find the initial velocity and use tan^-1 to find the angle, which will be ... degrees above the horizontal ground.

The answer I got for the initial velocity was 42m/s and the angle approximately 41 degrees.