Question

a ball is thrown upwards with a velocity of 10 meters per second what is the maximum height reached by the ball and the time taken for the ball to come back to the thrower?

Answer 1

Answer:

a= -10 m/s^2 [ a is acceleration due to gravity]

u= 10 m/s

v=0

using equation of motion,we have,

v^2-u^2=2as

0-100=2×(-10)×s

s=100/20

s=5m

now ,v=u+at

0= 10+(-10)t

-10= -10t

t= 1sec.

hope it helps you

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Answer 2

GiveN :

• Initial velocity (u) = 10 m/s
• Final Velocity (v) = 0 m/s

To FinD :

• Maximum height
• Time taken by ball to come back to thrower

SolutioN :

Use 3rd equation of motion :

$\dashrightarrow \large \boxed{\tt{v^2 \: - \: u^2 \: = \: 2as}} \\ \\ \dashrightarrow {\tt{0^2 \: - \: 10^2 \: = \: 2 \: \times \: -10 \: \times \: s}} \\ \\ \tt{-100 \: = \: -20s} \\ \\ \dashrightarrow \tt{100 \: = \: 20s} \\ \\ \dashrightarrow \tt{s \: = \: \dfrac{100}{20}} \\ \\ \dashrightarrow \tt{s \: = \: 5} \\ \\ \underline{\boxed{\sf{Maximum \: Height \: is \: 5 \: m}}}$

$\rule{150}{1}$

Use 1st equation of motion :

$\dashrightarrow \large {\boxed{\tt{v \: = \: u \: + \: at}}} \\ \\ \dashrightarrow \tt{0 \: = \: 10 \: + \: (-10)t} \\ \\ \dashrightarrow \tt{-10 \: = \: -10t} \\ \\ \dashrightarrow \tt{t \: = \: \dfrac{10}{10}} \\ \\ \dashrightarrow \tt{t \: = \: 1 \: s}$

✯ This is time taken by ball to reach Maximum Height.

For calculating the time interval of ball to come back to the thrower will be 1 * 2 = 2s