Question

A soccer ball is kicked so that at the peak of its trajectory it has a horizontal speed of 26 m/s, is 5.3 meters above the ground. The entire flight takes 2.5 seconds. How far away from the kicker does the soccer ball land?

Answer 1

Explanation:

An interesting and sometimes fun way to learn physics involves relating physics problems to the real world. In the case of finding projectile motion one can apply a real life scenario of someone kicking a soccer ball.

If a soccer ball is kicked so that at the peak of its trajectory it has a horizontal speed of 21 m/s, is six meters above the ground, and the entire flight takes 2.2 seconds the soccer ball will land 46.2 meters away from the kicker. Using real life scenarios to demonstrate otherwise difficult mathematical problems can make them both easier and more fun to deal with.

Answer 2

The soccer ball landed 65 m far way from the kicker.

Given:

The horizontal speed= uₓ= 26 m/s

The highest peak of the trajectory=$h_{max}$ = 5.3 m

Time taken by the entire flight= T= 2.5 s

To find:

The distance between the kicker and the soccer ball when it landed (R).

Solution:

We know that,

The horizontal speed= uₓ= ucosθ where u is the initial velocity and θ is the angle between projectile and x- axis.

uₓ= ucosθ

Put the value of uₓ in the above formula.

26= ucosθ

u= 26/cosθ                    ...............................(1)

We know that,

Time taken by the entire flight= T= 2usinθ/g where u is the initial velocity and θ is the angle between projectile and x- axis and g is the gravitational acceleration.

T= 2usinθ/g

Put the value of T, u ( from equation 1 ) and g (9.8 m/s²) in the above formula.

2.5= 2(26/cosθ )sinθ/9.8

2.5= 2(26)tanθ/9.8

tanθ= $\frac{(2.5)(9.8)}{(2)(26)}$                          .............................(2)

We know that,

The distance covered by the projectile= Range of the projectile= R= (usinθ)²/g where u is the initial velocity and θ is the angle between projectile and x- axis and g is the gravitational acceleration.

R= 2u²sinθcosθ/g

Put the values of u ( from equation 1 ) and g (9.8 m/s²) in the above formula.

R= 2(26/cosθ )²sinθcosθ/9.8

R= 2(26)²tanθ/9.8

Put the value of tanθ from equation 2.

R= $\frac{(2)(26)(26)(2.5)(9.8)}{(2)(26)(9.8)}$

R= 26×2.5

R= 65 m

Therefore, the soccer ball landed 65 m far way from the kicker.

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