Question

A soccer ball is kicked from the ground with an initial velocity of 25m/s at an...

A soccer ball is kicked from the ground with an initial velocity of 25m/s at an angle of 40 degrees with the horizontal. Determine how long the ball will stay in the air, before hitting the ground.

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Answer #1

Horizontal velocity= 25cos40

vertical velocity=25 sin40

applying s=ut+0.5at2 in vertical direction

0=(25sin40)t+0.5(-g)t2

t=(25sin40)/0.5g=3.2795sec

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Answer #2

SOLUTION 


Vertical component of initial velocity = uV = 25 sin(40) = 17.67767 m/s


Now, as ball goes up against gravity, velocity decreases at the rate of g = 9.8 m/s^2

=> v = uV - 9.8 t

At maximum height, velocity = 0

=> 0 = 17.67767 - 9.8 t 

=> t = time to reach maximum height = 17.67767/9.8 = 1.80 sec.


Time to come down to ground = time taken to reach maximum height = 1.80 sec


So,


Total time ball is in air

 = time to go up + time to come down 

= 1.80 + 1.80

= 3.60 sec. (ANSWER).

answered by: Tulsiram Garg

> There is a calculation error, 25 sin(40) = 16.07 m/s and not 17.67767. So final answer would be 3.28 sec and not 3.60 sec. Please see the correct answer in the next answer by me.

Tulsiram Garg Fri, Nov 26, 2021 7:50 AM

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Answer #3

SOLUTION 


Vertical component of initial velocity = uV = 25 sin(40) = 16.07 m/s


Now, as ball goes up against gravity, velocity decreases at the rate of g = 9.8 m/s^2

=> v = uV - 9.8 t

At maximum height, velocity = 0

=> 0 = 16.07 - 9.8 t 

=> t = time to reach maximum height = 16.07/9.8 = 1.64sec.


Time to come down to ground = time taken to reach maximum height = 1.64 sec


So,


Total time ball is in air

 = time to go up + time to come down 

= 1.64 + 1.64

= 3.28 sec. (ANSWER).

answered by: Tulsiram Garg
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