Physics, Motion of a Projectile.?

Question

A ball is thown from a point 1.0 m above the ground. the initial velociy is 19.6 m/s at an angle of 30.0 above the horizontal.

What is the maximum height of the ball above the ground?
What is the speed of the ball at its highest point in the tragectory?

Answer 1

First find the vertical and horizontal components of the velocity:

Vx = 19.6 cos 30 = 16.97 m/s
Vy = 19.6 sin 30 = 9.8 m/s

The maximum height will occur when the vertical velocity is zero.

Vf = Vo - gt
0 = 9.8 - 9.8t
t = 1

The height at this time is given by 1 + 0.5 gt^2 = 1 + 4.9 = 5.9 m

The speed of the ball at it's highest point is simply Vx = 16.97 m/s

Answer 2

x = xo + voxt = xo + vocos(30)t
y = y0 + voyt - 1/2 gt^2 = y0 + vosin(30)t - 1/2 gt^2, and the

y = 4.9 m

But vfy = voy -gt = vosin(30) - gt

In the maximum heihgt, vfy = 0, then

0 = 19.6 sin (30) - gt, 19.6(0.5)=gt.....t = (19.6 * 0.5 )/9.8 = 1s.

put this in y = 0 + 19.6 cos (30) t - 1/2gt^2.

to the second questiion, is :

vfy= 19.6 * 0.5 - 9.8 (1) = 0... obviously

but, vfx = x0 + voxt = 0 + 19.6 cos (30) * 1s = 16.97 m/s