Question

a ball is thrown upward from the ground at the tower with speed of 20m/s there is a window in the tower at the height of 15m from the ground. how many times and when will the ball pass the window? (g=10m/s^2)

Answer 1

Explanation:

Given A ball is thrown upward from the ground at the tower with speed of 20 m/s there is a window in the tower at the height of 15 m from the ground. how many times and when will the ball pass the window? (g=10 m/s^2)    

Initial velocity u = 20 m / s  

Maximum height ball will reach h = ?

We know that equation of motion is given by v^2 = u^2 + 2gh

(Therefore at maximum height v = 0)

So h = - u^2 / 2g

       = -(20)^2 / 2 ( 10)

      = 400 / 20

        = 20 m

So the ball reaches a height of 20 m and returns back. It passes window twice. Now we need to calculate time taken by ball to reach 15 m height.

So h = ut – 1/2 gt^2

    15 = 20 t – 1/2 (10)t^2

    5t^2 – 20 t + 15 = 0

So t^2 – 4t + 3 = 0

     t^2 – 3t – t + 3= 0

      t(t – 3) – 1(t – 3) = 0

     (t – 3) (t – 1) = 0

    Or t = 3,1

Therefore the ball will pass the window at 1 and 3 secs respectively.

Answer 2

Answer:

t=1 and t=3 seconds

Explanation:

Answer: t=1 and t=3 seconds

Explanation:

Initial velocity,u=20m/s

acceleration,a=-g

Height,S=15m

time,t=?

we know that

Substituting the values

Substituting g=10m/s²

So the body passes through the window at t=1 and t=3 seconds