First slide

Projection Under uniform Acceleration

Question

A boy throws a ball upwards with velocity v₀ = 20 m/s. The wind imparts a horizontal acceleration of 4 m/s² to the left. The angle $θ$ with the vertical at which the ball must be thrown, so that the ball returns to the boy’s hand is : (g = 10 m/s²)

Moderate

A

tan^-1 (1.2)
B

tan^-1 (0.2)
C

tan^-1 (2)
D

tan^-1 (0.4)

Solution

$\begin{array}{l} v_{y} = v_{0} \cos θ = 20 \cos θ \\ v_{x} = v_{0} \sin θ = 20 \sin θ \end{array}$
Time of flight of the ball is :
$T = \frac{2 v_{y}}{g} = \frac{40 \cos θ}{10} = 4 \cos θ . . . (i)$
In this time displacement of ball in horizontal direction should also be zero,
$i . e ., 0 = v_{x} T - \frac{1}{2} a_{x} T^{2}$
This gives, $T = \frac{2 v_{x}}{a_{x}} = \frac{(20 \sin θ) 2}{4}$
= 10 sin $θ$ .....(ii)
From eqn. (i) and eqn. (ii)
4 cos $θ$ = 10 sin $θ$
$∴ \tan θ = \frac{4}{10} = 0 .4$

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