First slide

Projectile motion

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

1s
B

2s
C

3s
D

4s

Solution

Let at any time t, the ball be at height of 15 m.
$\begin{array}{l} S_{y} = u_{y} t + \frac{1}{2} a_{y} t^{2} \\ \Rightarrow 15 = u \sin α t - \frac{1}{2} g t^{2} \end{array}$
$\begin{array}{l} \Rightarrow 15 = 52 \times \frac{5}{13} t - \frac{1}{2} \times 10 t^{2} \\ \Rightarrow t^{2} - 4 t + 3 = 0 \Rightarrow (t - 1) (t - 3) = 0 \\ \Rightarrow t = 1 s, t = 3 s . Required time is 3 - 1 = 2 s \end{array}$

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