Question

In one contest at the county fair, a spring-loaded plunger launches a ball at a speed of 3.2m/s from one corner of a smooth, flat board that is tilted up at a 20 degree angle. To win, you must make the ball hit a small target at the adjacent corner, 2.40m away. At what angle theta should you tilt the ball launcher?

Answer 1

Answer:

Explanation:

Given

Speed of ball $u=3.2\ m/s$

Plane is inclined at an angle $20^{\circ}$

To win the Game we need to hit the target at $x=2.4\ m$ away

Launch angle of ball $\theta$

Motion of ball can be considered in two planes i.e. Vertical to the plane and horizontal to the plane

So Net acceleration in vertical plane is $g\sin 20$

Range of Projectile is given by

$R=\frac{u^2\sin 2\theta }{g}$

for $R=2.4\ m$

$2.4=\frac{3.2^2\times sin 2\theta }{g\sin 20}$

$\sin 2\theta =\frac{2.4\times 9.8\times \sin 20}{3.2^2}$

$\sin 2\theta =0.7855$

$2\theta =51.77$

$\theta =25.88^{\circ}$

so ball must be launched at an angle of $25.88^{\circ}$