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(a) In order to determine the time at which the ball reaches its maximum height, use the following formula:
[tex]v_y=v_{oy}-gt[/tex]where,
vy: final speed at the maximum height = 0m/s
voy: vertical component of the initial velocity
g: gravitational acceleration constant = 9.8m/s^2
t: time
The vertical component of the initial velocity is:
[tex]v_{oy}=v_o\sin (53)=(\frac{20m}{s})(\sin (53))=\frac{15.97m}{s}[/tex]Solve the equation for t and replace the values of the rest of the parameters:
[tex]t=\frac{v_{oy}-v_y}{g}=\frac{\frac{15.97m}{s}-\frac{0m}{s}}{\frac{9.8m}{s^2}}=1.63s[/tex]Hence, after 1.63s the ball has reached its maximum height.
(ii) Use the following formula to determine the greatest height:
[tex]y_{\max }=\frac{v^2_{oy}}{2g}=\frac{(\frac{15.97m}{s})^2}{2\cdot\frac{9.8m}{s^2}}=13.01m[/tex]Hence, the maximum height is approximately 13.01m