Question

1. A ball is thrown upward with 20m/s from top of a building of height 60m.Find time taken by the ball to reach ground.

2. A ball thrown with certain velocity upward reaches a maximum height H. If
the same ball is thrown with triple velocity on a planet with half gravity as
that of earth. Find maximum height.

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Answer 1

Answer:

1. 2+2=4 sec

2.h=u²

Explanation:

u 20 , v 0 , g -10 , h 60 , t ??

1.v=u+gt

0=20-10t

10t=20

t=20/10=2

total t= 2t=2*2=4

Answer 2

$\small{ \bold { \blue{ question \: - 1}}}$

A ball is thrown upward with 20m/s from top of a building of height 60m.Find time taken by the ball to reach ground.

$\small{ \bold{ \green{answer :}}}$

Time to reach maximum height can be obtained from v = u + at

0 = 20 + ( - 10 )t

t = 2s

$s = \: ut + {0.5at}^{2} = 20(2) + 0.5( - 10) {(2)}^{2} = 20cm$

thus , total distance of maximum height is 45 m

$s \: = ut \: + {0.5at}^{2}$

$45 = 0 + 0.5(10) {(t)}^{2}$

t' = 3s

total time = 3 + 2 = 5s

$\small{ \bold{ \pink{question - 2}}}$

A ball thrown with certain velocity upward reaches a maximum height H. if the same ball is thrown with triple velocity on a planet with half gravity as that of earth . find maximum height .

$\small{ \bold { \purple{answer : }}}$

maximum height means final velocity is zero

$thus \: h \: = \frac{0 - {u}^{2} }{ - 2g} = \frac{ {u}^{2} }{2g}$

to triple maximum height , i .e . 3H we have

$3h = { \frac{u}{2g} }^{2} or \: {u}^{2} = {3u}^{2} \: or \: \: u \: = \sqrt{3}$