Question

A ball A is dropped from a 45 m high cliff. Two seconds later another ball B is thrown downwards from the same place with some initial speed. Two balls reach the ground together. Find the speed with which the ball B was thrown. (Take g = 10 ms– 2)

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Initial\:velocity\:of\:ball\:B=40\:m/s}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given :}} \\ \tt: \implies Height \: of \: cliff = 45 \: m \\ \\ \red{\underline \bold{To \: Find :}} \\ \tt: \implies Initial \: speed \: of \: ball \:B=?$

• According to given question :

$\tt \circ \: Initial \: velocity = 0 \: m/s \\ \\ \tt \circ \: Acceleration = {10 \: m/s}^{2} \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} {at}^{2} \\ \\ \tt: \implies 45 = 0 \times t + \frac{1}{2} \times 10 \times {t}^{2} \\ \\ \tt: \implies 45 \times 2 = {10 \times t}^{2} \\ \\ \tt: \implies \frac{90}{10} = {t}^{2} \\ \\ \tt: \implies {t}^{2} = 9 \\ \\ \green{\tt: \implies t = 3 \: sec} \\ \\ \tt \circ \: Time \: taken = 3 - 2 = 1 \: sec \\ \\ \tt \circ \: Let \:Initial \: velocity \: be \: 'u' \\ \\ \bold{As \: we \: know \: that} \\ \tt: \implies s = ut + \frac{1}{2} {at}^{2} \\ \\ \tt: \implies 45 = u \times 1 + \frac{1}{2} \times 10 \times {1}^{2} \\ \\ \tt: \implies 45 = u + 5 \\ \\ \tt: \implies u = 45 - 5 \\ \\ \green{\tt: \implies u = 40 \: m/s} \\ \\ \green{\tt \therefore Initial \: velocity \: of \: ball \: b\: is \: 40 \: m/s}$

Answer 2

Answer:-

$\sf{The \ initial \ velocity \ of \ ball \ B \ is}$

$\sf{40 \ m \ s^{-1}}$

Given:

• A ball A is dropped from a 45 m high cliff. Two second later another ball B is thrown downwards from the same place with some initial speed.

• Two balls reach the ground together.

To find:

• The speed with which the ball B was thrown.

Solution:

$\sf{For \ ball \ A,}$

$\sf{Initial \ velocity(u)=0,}$

$\sf{Distance(s)=45 \ m,}$

$\sf{Acceleration \ due \ to \ gravity(a)=10 \ m \ s^{-2}}$

$\sf{According \ to \ the \ given \ condition.}$

$\sf{The \ ball \ A \ is \ in \ free \ fall.}$

$\sf{Kinematic \ equations:-}$

$\sf{v=u+at}$

$\sf{s=ut+\frac{1}{2}\times \ at^{2}}$

$\sf{v^{2}=u^{2}+2as}$

$\sf{According \ to \ the \ second \ equation \ of \ motion.}$

$\sf{s=ut+\frac{1}{2}\times \ at^{2}}$

$\sf{\therefore{45=0+\frac{1}{2}\times10\times \ t^{2}}}$

$\sf{\therefore{t^{2}=\frac{45}{5}}}$

$\sf{\therefore{t^{2}=9}}$

$\sf{\underline{\underline{\therefore{t=3 \ s}}}}$

$\sf{For \ ball \ B,}$

$\sf{Time(t)=3-2=1 \ second,}$

$\sf{Distance (s)=45 \ m,}$

$\sf{Acceleration \ due \ to \ gravity(a)=10 \ m \ s^{-2}}$

$\sf{But, \ ball \ B \ has \ some \ initial \ velocity(u).}$

$\sf{According \ to \ the \ second \ equation \ of \ motion.}$

$\sf{s=ut+\frac{1}{2}\times \ at^{2}}$

$\sf{\therefore{45=u(1)+\frac{1}{2}\times(10)(1^{2})}}$

$\sf{\therefore{45=u+5}}$

$\sf{\therefore{u=45-5}}$

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

$\sf\purple{\tt{\therefore{The \ initial \ velocity \ of \ ball \ B \ is}}}$

$\sf\purple{\tt{40 \ m \ s^{-1}}}$