First slide

Projection Under uniform Acceleration

Question

A golfer standing on level ground hits a ball with a velocity of u = 52 ms^-1 at an angle $α$ above the horizontal. If tan $α$ = 5/12, then the time for which the ball is at least 15 m above the ground will be (take g = 10 ms-²)

Moderate

A

1 s
B

2 s
C

3 s
D

4 s

Solution

Let at any time t, the ball be at height of 15 m.
$S_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2} 15 = usinαt - \frac{1}{2} {gt}^{2} 15 = 52 \times \frac{5}{13} t - \frac{1}{2} \times 10 t^{2} t^{2} - 4 t + 3 = 0 (t - 1) (t - 3) = 0 t = 1 s, t = 3 s . Required time is 3 - 1 = 2 s$

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