How long does it take to reach its highest point? Answer in units of s. What is velocity when it returns to the level from which it started? ans. m/s
A ball is thrown vertically upward with a speed of 26.3 m/s.
The equation for the velocity for the ball thrown vertically :
v= u+at, where u is the initial velocity a is the retardartion due gravitional acceleration, and t is the time of the motion.
At the highest point the velocity v of the ball is zero.
Therefore, 0=26.3-9.8t or 9.8t=26.3. Therefore, t= 26.3/9.8=2.6837 is the time to reach the heighest point.
The height reached s= ut-(1/2)gt^2 = 26.3(26837)-(1/2)(9.8)(2.6837^2)=35.2903m
Returning tothe ground:
Initial velocity u= 0, final velocity v to be found and acceleraion is the acceleration due to gravitation.The height s=35.2903m
We know that v^2-u^2= 2as
v= sqrt(2as)= sqrt(2*9.8*35.2903)=26.3 m/s towards the earth. It is equival in magnitude to the starting velocity but opposite in direction.
The ball reaches its highest point when the velocity becomes 0 under the influence of gravity.
Therefore time taken to reach highest point = (initial velocity)/acceleration
= 26.3/9.81 = 2.6809 seconds.
The velocity when it returns to the original is point is exactly equal and opposite of initial velocity.
Thus velocity when ball returns to starting point = -26.3 m/s