Physics, asked by ashrithdevara9174, 1 year ago

A man throws ball with the same speed vertically upward one after another at the interval of 2 second what should be the speed of the throw so that more than balls are in the sky at the time

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Answered by Anonymous
0

Here is your solution buddy ↙️↙️☺️☺️

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

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

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.

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.

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

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

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

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

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

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^2u= 9.8 m/s

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^2u= 9.8 m/s

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^2u= 9.8 m/s speed has to be more than 9.8 m/s

Hope this helps you ✌️✌️☺️☺️❤️❤️

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