Answer:
a) v= 0 m/s b) v= 6.86 m/s
Explanation:
a) When the ball reaches to its highest point, under the influence of gravity, before starting to fall down, it momentarily comes to an stop (this is needed prior to change direction in any movement), so, applying the definition of acceleration, and replacing the acceleration a by g, we have:
vf = v₀ - g*t (1)
The minus sign means that the acceleration due to gravity is always downward, so if we assume that the positive direction is upwards it must be negative.
At the highest point, vf= 0.
b) Prior to solve this point, we need to know which is the time when the ball reaches to its highest point.
As we know vf=0, we can solve (1) for t, as follows:
th = v₀/g
Now, for a time that is 0.7 s before this time, applying the acceleration definition and solving for v again, we have:
v = v₀ -(g *(th-0.7 s)), but th= v₀/g, so we get:
v= v₀ -g((v₀/g)-0.7 s) = v₀ - v₀ + g*0.7 s
⇒ v=g*0.7 s = 9.8 m/s²*0.7 s
⇒ v = 6.86 m/s
.
, so 
. We use 
and then the final speed must be 0 because it stops at the highest point. So 
. Solve for 
and you get 
, and then we plug the values: 
and we already have the time from "b)", so ![Y_m_a_x = [(32sin(25))*(32sin(25)/10)] - 5(32sin(25)/10)^2](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%20%5B%2832sin%2825%29%29%2A%2832sin%2825%29%2F10%29%5D%20-%205%2832sin%2825%29%2F10%29%5E2)
; then we just rearrange it ![Y_m_a_x = 10[(32sin(25))^2/100] - 5 [(32sin(25))^2/100]](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%2010%5B%2832sin%2825%29%29%5E2%2F100%5D%20-%205%20%5B%2832sin%2825%29%29%5E2%2F100%5D%20)
and finally ![Y_m_a_x = 5[(32sin(25))^2/100] = (32sin(25))^2/20](https://tex.z-dn.net/?f=Y_m_a_x%20%3D%205%5B%2832sin%2825%29%29%5E2%2F100%5D%20%3D%20%2832sin%2825%29%29%5E2%2F20)