Question

a ball is filed with an initial velocity vo=36i m/s+48j m/s from the ground g=10m/s. what time taken the highest point.​

Answer 1

Given,

Initial velocity of the ball, vo = 36i m/s +48j m/s

The magnitude of the initial velocity, |vo|=$\sqrt{(36^2+48^2)} = 60$ m/s

acceleration due to gravity, g = 10 $m/s^2$

To Find,

Time taken by the ball to reach the highest point, t =?

Solution,

At the highest point, the ball will have velocity, v=0 m/s and acceleration g towards the center of the earth.

We know, v=u+gt

where, v=final velocity=0 m/s

u=initial velocity=vo=60 m/s

a=acceleration=g= -10 $m/s^2$

(the negative sign implies that the acceleration is applied in the opposite side of the motion)

Therefore, 0=60+(-10*t)

t= 6 sec.

Hence, the time taken to reach the highest point is 6 sec.