Question

A football is kicked from the ground level at an angle of 55 degrees, with respect to the horizontal. With an initial speed of 15 m/s. What is the horizontal displacement at the end of 3 seconds.

And how long does it take the football to reach the highest point in trajectory?

What is the approximate magnitude of the vertical component of the final velocity of the football as it reaches the ground?

How long is the football in the air before striking the ground?

Answer 1

This can be considered as a projectile motion.

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PART 1:

Initial horizontal velocity, V_x = V.cosθ = (15 m/s) x cos 55^o = 8.6 m/s

Time taken, t = 3 s

Therefore, Horizontal displacement = V_xt = (8.6 m/s) x (3 s) = 25.8 m

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PART 2:

We need to consider only the vertical motion of the projectile.

Initial vertical velocity, V_y = V.sinθ = (15 m/s) x sin 55^o = 12.287 m/s

Acceleration, g = -9.8 m/s²

At the maximum height, final vertical speed of the projectile will be zero.

Therefore, Time taken to reach maximum height is given by:

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PART 3:

We can neglect the air resistance.

In that case, the final vertical speed of the projectile (when hitting the ground) will be equal to the initial vertical speed.

Therefore, magnitude of the vertical component of the final velocity

of the football as it reaches the ground = V_y = 12.287 m/s

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PART 4:

Time of flight of a projectile is given by:

(or you can also follow this method, T = 2t = 2 x 1.25 s = 2.51 s) where t is the time required to reach maximum height, or we can call it as time of ascend.

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​​​​​​​(Please rate the answer if you are satisfied. In case of any queries, please reach out to me via comments)