A-man-is-standing-on-top-of-a-building-100-m-high-He-throws-two-balls-vertically-one-at-t-0-and-other-after-a-time-interval-less-than-2-seconds-The-later-ball-is-thrown-at-a-velocity-of-half-th

Question Number 17816 by Tinkutara last updated on 11/Jul/17

A man is standing on top of a building

100 m high . He throws two balls

vertically, one at t = 0 and other after

a time interval (less than 2 seconds) .

The later ball is thrown at a velocity of

half the first . The vertical gap between

first and second ball is + 15 m at t = 2 s .

The gap is found to remain constant .

Calculate the velocity with which the

balls were thrown and the exact time

interval between their throw .

Answered by mrW1 last updated on 11/Jul/17

h_{1} = v_{1} t + \frac{1}{2} {gt}^{2}

h_{2} = \frac{1}{2} v_{1} (t - Δ t) + \frac{1}{2} g {(t - Δ t)}^{2}

Δ h = h_{1} - h_{2} = \frac{1}{2} v_{1} (t + Δ t) + \frac{1}{2} g (2 t - Δ t) Δ t

= \frac{1}{2} v_{1} t + gt Δ t + \frac{1}{2} v_{1} Δ t - \frac{1}{2} g Δ t^{2}

= t (\frac{1}{2} v_{1} + g Δ t) + \frac{1}{2} Δ t (v_{1} - g Δ t)

since Δ h remains constant

\Rightarrow \frac{1}{2} v_{1} + g Δ t = 0

\Rightarrow v_{1} = - 2 g Δ t

Δ h = \frac{1}{2} Δ t (v_{1} - g Δ t) = - \frac{3}{2} g Δ t

- 15 = - \frac{3}{2} g Δ t

\Rightarrow Δ t = 1 s

\Rightarrow v_{1} = - 2 g \times 1 = - 20 m / s (⇈)

Commented by Tinkutara last updated on 11/Jul/17

Thanks Sir!

Answered by ajfour last updated on 11/Jul/17

Assuming, balls are thrown

vertically upwards . First ball at a

velocity 2 u, and second after time

t_{0} at velocity u .

if t > t_{0} we have

s_{1} - s_{2} = 15 (s_{1} - s_{2} = - 15 not possible)

[2 ut - \frac{1}{2} {gt}^{2}] - [u (t - t_{0}) - \frac{1}{2} g {(t - t_{0})}^{2}] = 15

\Rightarrow u (t + t_{0}) - \frac{g}{2} t_{0} (2 t - t_{0}) - 15 = 0

since this is true for all times t > t_{0}

until any ball hits the ground, so

coefficient of t in equation and the

constant term are both zero .

u - {gt}_{0} = 0 and {ut}_{0} + \frac{{gt}_{0}^{2}}{2} - 15 = 0

or u = {gt}_{0} and substituting this in

second equation, we get

\frac{{3 gt}_{0}^{2}}{2} = 15 \Rightarrow t_{0} = \sqrt{\frac{10}{g}}

u = {gt}_{0} = \sqrt{10 g}

if g = 10 m / s^{2}, t_{0} = 1 s and u = 10 m / s .

Commented by Tinkutara last updated on 11/Jul/17

Thanks Sir!