Question

A boy throws a ball upward with a speed v0 = 12 m/s. The wind imparts a horizontal acceleration of 0.4 m/s2 to the left. At what angle θ must the ball be thrown so that it returns to the point of release? Assume that the wind does not affect the vertical motion.

Answer 1

Answer:

The angle is 2.33°.

Explanation:

Given that,

Speed of ball = 12 m/s

Acceleration = 0.4 m/s²

We need to calculate the time

Using formula of time of flight

$t=\dfrac{2u}{g}$

$t=\dfrac{2v\cos\theta}{g}$

Put the value into the formula

$t=\dfrac{2\times12\cos\theta}{9.8}$

$t=2.44\cos\theta$

We need to calculate the angle

Using equation of motion along vertical direction

$s=ut-\dfrac{1}{2}at^2$

$s=v\sin\theta\times t-\dfrac{1}{2}at^2$

Put the value in the equation

$0=12\sin\theta\times2.44\cos\theta-\dfrac{1}{2}\times0.4\times(2.44\cos\theta)^2$

$2\times12\sin\theta\times2.44=0.4\times(2.44)^2\cos\theta$

$\tan\theta=\dfrac{0.4\times2.44}{2\times12}$

$\theta=\tan^{-1}(0.04066)$

$\theta=2.33^{\circ}$

Hence, The angle is 2.33°.