Question

a ball is thrown up vertically returns to the thrower after 4s find the maximum height it reaches and it's position after 3s​

Answer 1

Given:

A ball is thrown up vertically returns to the thrower after 4s .

To find:

• Max height reached ?

• Position of the ball after 3 seconds of throwing ?

Calculation:

If the ball takes 4 seconds to return back to the thrower (we know that the time of ascent is equal to the time of descent) , so the time to reach ground from max height is 4/2 = 2 seconds.

Let that displacement be h :

$\therefore \: h \: = ut \: + \dfrac{1}{2} g {t}^{2}$

• When ball is coming down from Max height, its initial velocity is zero.

$\implies \: h \: = 0 \: + \dfrac{1}{2} g {t}^{2}$

$\implies \: h \: = \dfrac{1}{2} \times 10 \times {(2)}^{2}$

$\implies \: h \: = 20 \: m$

So, the max height is 20 metres.

Now, position at 3 seconds after throwing is equivalent to position at (3-2) = 1 second after starting from max height.

$\therefore \: d \: = ut \: + \dfrac{1}{2} g {t}^{2}$

$\implies \: d \: = 0\: + \dfrac{1}{2} g {t}^{2}$

$\implies \: d \: = \dfrac{1}{2} g {t}^{2}$

$\implies \: d \: = \dfrac{1}{2} g {(1)}^{2}$

$\implies \: d \: = \dfrac{1}{2} \times 10$

$\implies \: d \: = 5 \: metres$

So, the balls position after 3 seconds of throwing will be 5 metres from max height or (20-5) = 15 metres from ground.