Question

A ball thrown vertically upward return back studio thrower in 4 seconds total distance covered by the ball is take g is equal to 10 metre per second square

Answer 1

Answer:

initial velocity - 20m

Explanation:

s=ut+1/2 at^2

= u×4+1/2 (-g)4^2

= 4u - 10×8

= 4 u- 80

s = u- 20m

Answer 2

$d=160\ m$ is the distance covered by the ball.

Explanation:

Considering the half distance of the motion of the body when it is thrown from the ground till it is at the top height at rest before returning down the motion.

Given:

• initial velocity, $u=?$

• final velocity, $v=0\ m.s^{-1}$

• time taken, $t=4\ s$

• acceleration due to gravity, $g= 10\ m.s^{2}$

Using equation of motion:

$v=u-g.t$

$0=u-10\times 4$

$u=40\ m.s^{-1}$

Now,

$h=u.t-\frac{1}{2} g.t^2$

$h=40\times 4-\frac{1}{2}\times 10\times 4^2$

$h=80\ m$

Therefore total distance on returning down from top will be:

$d=2h$

$d=160\ m$

TOPIC: equation of motion under gravity

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