Answer to Question #206822 in Mechanical Engineering for Rodel

Question

A ball was thrown outside the window at an angle of 20° below the horizontal with an initial velocity of 8 m/s. If it lands 3 sec later, how far from the base of the building did it land?

Answer 1

Let ball is thrown with initial velocity 'V'=8 m/s

Since, it is launched at angle 20 degrees below horizontal,

The initial horizontal velocity

Vx= Vcos(20)=8cos(20)=7.52 m/s

The initial vertical velocity

Vy= -Vsin(20)=-2.74 m/s

In the horizontal direction velocity constant,

Horizontal distance=(Vx)×(Time)

Horizontal distance=(7.52)×(3)=22.55 m

Hence ball lands at distance 22.55m from base.