Question

the ball is. thrown vertically upward and returns 8 seconds later find the greatest height reached by the ball and intial velocity of ball​

Answer 1

Answer:

the time for upward journey equals time for downward journey. That is time taken by the ball to go up = time taken by the ball to fall down from the highest point = 8/2 seconds = 4 seconds

When the ball begins to fall from the highest point its initial velocity u =0 m/s. It is falling under the action of acceleration due to gravity and g = acceleration due to gravity = - 10 m/s² (negative sign as g is directed downwards). The displacement in 4 seconds can be obtained using the formula :

s = u t + ½ a t², where u= 0 m/s, a = g =-10 m/s² and t = 4 seconds.

s = ½ × -10 × 4² = -80 m ( - sign signifies displacement is downwards)

The highest point is upwards in opposite direction to the displacement ie highest point = -s = +80m.

The highest point reached is 80 m above the throwing point.

Let the initial velocity with which the ball is thrown is + u m/s (positive sign signifies that u is directed up). The velocity becomes v = 0 m/s in time 4 seconds (the time of upward travel). Using

v = u + a t, we can get u remembering that a = g =-10m/s². Substituting we get;  0 = u - 10 × 4 s; ==> u = 40 m/s.

So the ball is thrown up with an upward initial velocity of 40 m/s and reaches the highest point at 80 m above the throwing point.