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.
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
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).
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).
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> 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