Question

A ball is thrown upwards at an unknown speed. in a time of

Answer 1

Answer:

a)  Initial speed of the ball = 14.45 m/s

b) At height 6 m speed of ball = 9.55 m/s

c) Maximum height reached = 10.65 m

Explanation:

a)  We have equation of motion $s=ut+\frac{1}{2} at^2$, where s is the displacement, u is the initial velocity, t is the time taken and a is the acceleration.

s = 6 m, t = 0.5 seconds, a = acceleration due to gravity value = -9.8$m/s^2$

 Substituting

    $6=u*0.5-\frac{1}{2} *9.8*0.5^2\\ \\ u=14.45m/s$

 Initial speed of the ball = 14.45 m/s

 b) We have equation of motion $v^2=u^2+2as$, where v is the final velocity

   s = 6 m, u = 14.45 m/s, a = -9.8$m/s^2$

    Substituting

        $v^2=14.45^2-2*9.8*6\\ \\ v=9.55m/s$

  So at height 6 m speed of ball = 9.55 m/s

c) We have equation of motion $v^2=u^2+2as$, where v is the final velocity

   u = 14.45 m/s, v =0 , a = -9.8$m/s^2$

   Substituting

     $0^2=14.45^2-2*9.8*s\\ \\ s=10.65 m$

  Maximum height reached = 10.65 m