Grade 11Mechanics

A man throws balls with the same speed vertically upwards one after the other at an interval of 2seconds.What should be the speed of throw so that more than two balls are in the sky at any time?

chakrala mounica

15 Years agoGrade 11

5 Answers

Rathod Shankar AskiitiansExpert-IITB

Approved Tutor Answer15 Years ago

Dear chakrala mounica

the first has to be above the ground when the second ball is thrown from the ground

so the first ball has to be above from the ground for 2 seonds.

s=ut-1/2*gt^2 t=2 and s=0 when first ball reaches the ground

0= 2*u- 1/2*g*4

u=g m/s take g=9.8 m/s^2

u= 9.8 m/s

speed has to be more than 9.8 m/s

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Rathod Shankar Singh

IIT Bombay

Raunak Chaudhury

9 Years ago

S=ut-1/2gt×t S=2u-19.6v^2=u^2-2aS0^2=u^2-2×9.8×(2u-19.6)0=u^2- 39.2u - (19.6)^20=(u- 19.6)^2u=19.6 m/secMore than 2 balls speed is greater than 19.6 m/sec

sahil

8 Years ago

more than 19.6 m/s Let the required speed of through be u m/s. The time taken to reach maximum height t=u/g ; For two balls to remain in air at any time, t must be greater than 2. ; ∴ u/g >2 ⇒ u > 2(g) ; u > 19.6 m/s

vansh tayal

7 Years ago

For at least 3 balls to remain above the ground, the time of flight of the first ball must be more than or equal to 4 s. Because in t = 4 s the third ball leaves the ground.

Now, for the first ball,

Initial velocity = u

Final velocity = -u

Acceleration = -g

So, -u = u – gt

=> u = gt/2

=> u = 9.8 × 4/2

=> u = 19.6 m/s

so the speed should be greater than 19.6 m/s to keep more than 2 balls in the sky at any time.

Thus, the ball must be thrown at speed more than or equal to 19.6 m/s u