Question

A ball is projected upwards from the ground with an initial velocity u the ball is at height of 160 m twice within the time interval of 4 second then initial speed us

Answer 1

A ball is projected upward from the ground with an initial velocity u , the ball is at height of 160m twice with the interval of 4sec.

here, S = 160m , a = g = -10 m/s² [ negative sign indicates acceleration acts against the motion of body]

use formula , S = ut + 1/2 at²

160 = ut - 1/2 × 10 t²

160 = ut - 5t²

5t² - ut + 160 = 0

t = [u ± √{u² - 4(5)(160)}]/2(5)

= [u ± √(u² - 3200)]/10

so, $t_1$ = {u + √(u² - 3200)}/10

$t_2$ = {u - √(u² - 3200)}/10

A/C to question,

$t_1-t_2=4$

⇒$\frac{u+\sqrt{u^2-3200}}{10}-\frac{u-\sqrt{u^2-3200}}{10}=4$

⇒$\frac{2\sqrt{u^2-3200}}{10}=4$

⇒$\frac{\sqrt{u^2-3200}}{10}=2$

⇒$\sqrt{u^2-3200}=20$

squaring both sides,

⇒u² - 3200 = 400

⇒u² = 3600 => u = ±60 m/s

we should ignore negative velocity.

hence, initial velocity , u = 60 m/s

Answer 2

Answer:

u = 60 m/s is your answer

Explanation: