Question

True or False (& Justify): If a ball is thrown horizontally from the top edge of a 150m building at a speed of 25 m/sec, then the ball’s velocity and speed 3 seconds later will be <25, -39.2> and 46.493 m/sec.

Answer 1

Answer:

38.6 m/s

Step-by-step explanation:

The motion of the ball is a projectile motion, which consists of two independent motions:  

- A uniform motion (constant velocity) along the horizontal direction  

- A uniformly accelerated motion, with constant acceleration (acceleration of gravity) in the downward direction  

Therefore we have to analyze the horizontal and vertical motion separately.

Along the horizontal direction, the velocity is constant during the motion, since there are no forces acting in this direction. So the horizontal velocity 3 seconds after the launch will be the same as the velocity at the launch:

$v_x = v_0 = 25 m/s$

The vertical velocity instead changes according to the suvat equation:

$v_y = u_y - gt$

where

$u_y=0$ is the initial vertical velocity

$g=9.8 m/s^2$ is the acceleration due to gravity

t is the time

Therefore, after t = 3 s,

$v_y=0-(9.8)(3)=-29.4 m/s$

So the velocity after 3 seconds is < 25, -29.4 > m/s. The magnitude of the velocity is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{25^2+(-29.4)^2}=38.6 m/s$