Question

if you toss a ball upward with a certain initial speed, it falls freely and reaches a maximum height 'h' a time t after it leaves your hand. If you throw the ball upward with double the initial speed what new maximum height does the ball reach? *

1 point

h√2

4h

8h

16h​

Answer 1

Given:

Ball thrown upwards reaches height h after time t.

To find:

Max height when initial velocity is doubled?

Calculation:

When initial velocity is u :

${v}^{2} = {u}^{2} - 2gh$

$\implies \: {0}^{2} = {u}^{2} - 2gh$

$\implies \: {u}^{2} = 2gh$

$\implies \: h = \dfrac{ {u}^{2} }{2g}$

When initial velocity is 2u:

${v}^{2} = {(2u)}^{2} - 2g(h2)$

$\implies \: {0}^{2} = 4{u}^{2} - 2g(h2)$

$\implies \: 4{u}^{2} = 2g(h2)$

$\implies \: h2 = 4 \times \dfrac{ {u}^{2} }{2g}$

$\implies \: h2= 4 h$

So, new height is 4h.