a) The acceleration is due to gravity at any given point if you look at it vertically, so [tex]-10 m/s^2[/tex].
b) [tex]sin(25) = V_y/V[/tex], so [tex]V_y = V*sin(25)[/tex]. We use [tex]V = V_0 + a t[/tex] and then the final speed must be 0 because it stops at the highest point. So [tex]0 = V_y - 10t[/tex]. Solve for [tex]t[/tex] and you get [tex]t = 32sin(25)/10 = 16sin(25)/5[/tex]
c) [tex]Y = Y_0 + V_0t + (1/2)at^2[/tex], and then we plug the values: [tex]Y_m_a_x = 32sin(25)*t - (1/2)*10*t^2[/tex] and we already have the time from "b)", so [tex]Y_m_a_x = [(32sin(25))*(32sin(25)/10)] - 5(32sin(25)/10)^2[/tex]; then we just rearrange it [tex]Y_m_a_x = 10[(32sin(25))^2/100] - 5 [(32sin(25))^2/100] [/tex] and finally [tex]Y_m_a_x = 5[(32sin(25))^2/100] = (32sin(25))^2/20[/tex]
