Answer
(a) The speed at the highest point is 7.1 m/s.
(b) The ball goes up to a height of 1.4 meters.
Work Step by Step
(a) The initial speed v is 8.8 m/s at an angle of $36^{\circ}$.
This means that the initial horizontal speed is:
$v_x = v~cos(\theta)$
$v_x = 8.8 ~m/s\times cos(36)$
$v_x = 7.1 ~m/s$
At the highest point, the speed will be $v_x$.
The speed at the highest point is 7.1 m/s.
(b) The initial kinetic energy will equal the kinetic energy plus the potential energy at the maximum height.:
$KE_2 + PE = KE_1$
$\frac{1}{2}mv_x^2 + mgh = \frac{1}{2}mv^2$
$h = \frac{v^2-v_x^2}{2g}$
$h = \frac{(8.8 ~m/s)^2-(7.1 ~m/s)^2}{2(9.8 ~m/s^2)} = 1.4 ~m$
The ball goes up to a height of 1.4 meters.
