Question

Very lost please help.

Answer 1

A) The acceleration is due to gravity at any given point if you look at it vertically, so $-10 m/s^2$.

b) $sin(25) = V_y/V$, so $V_y = V*sin(25)$. We use $V = V_0 + a t$ and then the final speed must be 0 because it stops at the highest point. So $0 = V_y - 10t$. Solve for $t$ and you get $t = 32sin(25)/10 = 16sin(25)/5$

c) $Y = Y_0 + V_0t + (1/2)at^2$, and then we plug the values: $Y_m_a_x = 32sin(25)*t - (1/2)*10*t^2$ and we already have the time from "b)", so $Y_m_a_x = [(32sin(25))*(32sin(25)/10)] - 5(32sin(25)/10)^2$; then we just rearrange it $Y_m_a_x = 10[(32sin(25))^2/100] - 5 [(32sin(25))^2/100]$ and finally $Y_m_a_x = 5[(32sin(25))^2/100] = (32sin(25))^2/20$