Gravity Velocity, max. height and gravity

Classical Mechanics Level 2

A ball is thrown straight up is caught back by the thrower after 6 sec. Calculate

1) The velocity with which the ball was thrown up,

2) The max. height it reaches and

3) The distance travelled by the ball for 2 seconds after it hits the maximum height.

DETAILS AND ASSUMPTIONS :

i) Give the answer in the form, a n s w e r 1 + a n s w e r 2 + a n s w e r 3 answer 1 + answer 2 + answer 3 .

ii) To make the calculations easier take the value of acceleration due to gravity as, g = 10 m / s 2 g = 10 m/s^2


The answer is 95.

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Sravanth C. by Sravanth C. , 21, India

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2 solutions

Chew-Seong Cheong
Mar 11, 2015

  1. Let the velocity with which the ball was thrown up be v 0 v_0 , then the velocity at time t t after it was thrown was: v ( t ) = v 0 − g t v(t) = v_0 - gt . Since the ball took 6 s 6s to came back to the thrower's hand, it took half that time, 3 s 3s , to reached the maximum height when its velocity is zero. ⇒ v ( 3 ) = v 0 − 10 ( 3 ) = 0 ⇒ v 0 = 30 m s − 1 \Rightarrow v(3) = v_0 - 10(3) = 0\quad \Rightarrow v_0 = 30ms^{-1} .

  2. Vertical distance on the way up was given by s ( t ) = v 0 t − 1 2 g t 2 s(t) = v_0 t - \frac {1}{2} g t^2 . Therefore, the maximum height h = s ( 3 ) = 30 ( 3 ) − 1 2 ( 10 ) ( 9 ) = 90 − 45 = 45 m h = s(3) = 30(3) - \frac {1}{2}(10)(9) = 90 - 45 = 45m .

  3. The distance when the ball was coming down was given by s ( t ) = 0 + 1 2 g t 2 ⇒ s ( 2 ) = 1 2 ( 10 ) ( 4 ) = 20 m s(t) = 0 + \frac {1}{2} g t^2\quad \Rightarrow s(2) = \frac{1}{2}(10)(4) = 20m .

Then, the required answer is 30 + 45 + 20 = 95 30+45+20 = \boxed{95}

11 Helpful 0 Interesting 0 Brilliant 0 Confused

Perfect again! you've done a wonderful job sir!!!

Please don't feel bad, Are you really 57 years old sir???

Sravanth C. - 6 years, 3 months ago

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Of course, Sravanth. And that's the Amazing!!!

Muhammad Arifur Rahman - 6 years, 1 month ago

Yes, I am 57 years old.

Chew-Seong Cheong - 6 years, 1 month ago

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Thanks for replying . . .

Sravanth C. - 6 years, 1 month ago

something is wrong with the question....should use displacement but not distance....question has problem

Zack Yeung - 6 years, 1 month ago

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yes even i think so .......they should mention whether we take the displacement from bottom or top and acc. to my calculations its 30+45+30 which is 115 instead of 95

roger potter - 5 years, 5 months ago
Anoorag Nayak
Nov 13, 2015

Isn't a similar question is in class 9 NCERT science textbook?

0 Helpful 0 Interesting 0 Brilliant 0 Confused

yes .....but the last part {dist. one is wrong.........i think]

roger potter - 5 years, 5 months ago

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