Question

A football player punts the ball at a 45.0º angle. without an effect from the wind, the ball would travel 60.0 m horizontally. (a) what is the initial speed of the ball? (b) when the ball is near its maximum height it experiences a brief gust of wind that reduces its horizontal velocity by 1.50 m/s. what distance does the ball travel horizontally?

Answer 1

By using projectile motion, the initial speed of the ball without any effect is √6 m/s and the ball will travel (60 - 15√3) m horizontally affected by wind.

We need to know projectile motion to solve this problem. The projectile motion is divided into 2 axes which x-axis and the y-axis.

Vx = Vcosθ

Vy = Vsinθ

Xmax = (V² × sin 2θ) / g

Ymax = v² sin² θ/2g

Apply the equation to uniform motion

V1 = V + a . t

s = V . t + 1/2 a . t²

V1² = V² + 2 a . s

From the question above, we know that:

θ = 45⁰

Xmax = 60m

g = 10 m/s²

By using terms of maximum horizontal distance of projectile motion, we get

Xmax = (V² × sin 2θ) / g

60 = V² × sin (2 . 45) / 10

V² = 6

V = √6 m/s

Hence, the initial speed of the ball without any effect is √6 m/s

The initial horizontal velocity is

Vx = √6 cos(45)

Vx = √3 m/s

Tmax = Xmax / Vx

Tmax = 60 / √3

Tmax = 20√3 second

Then, the time of reaching maximum height is

Tpeak = 1/2 x 20√3 = 10√3 second

Ball experience a brief gust of wind (Vw = 1.50 m/s). Then, the total horizontal distance is

Xtotal = Xup + Xdown

Xtotal = Vx . Tpeak + (Vx - Vw) . Tpeak

Xtotal = √3 . 10√3 + (√3 - 1.5) . 10√3

Xtotal = 30 + 30 - 15√3

Xtotal = (60 - 15√3) m

Hence, the ball will travel (60 - 15√3) m horizontally affected by wind.

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