A football is kicked, from ground level, at an angle of 45 degrees above horizon

Question

A football is kicked, from ground level, at an angle of 45 degrees above horizontal. It needs to clear 3.0 high crossbar, located 18m from where the ball is kicked. You may use sin(45) = cos(45) = 0.707.

a) With what minimum speed must the ball be kicked to clear the bar?

b) If kicked at the speed in part-a, what maximum height will it reach?

c) If kicked at the speed found in part-a, how many meters down the field will it be when it reaches its maximum height?

Please show all work and formulas used.

Explanation / Answer

Vo = initial speed of launch

consider the motion along the X-direction

Vox = initial velocity = Vo Cos45

X = displacement = 18 m

t = time taken

using the equation

X = Vox t

18 = (Vo Cos45 ) t

t = 18/(Vo Cos45 ) eq-1

consider the motion along the Y-direction

Voy = initial velocity = Vo Sin45

Y = vertical displacement = 3 m

a = acceleration = - 9.8

using the equation

Y = Voy t + (0.5) a t2

3 = (Vo Sin45) (18/(Vo Cos45 )) + (0.5) (-9.8) (18/(Vo Cos45 ))2

Vo = 14.5 m/s

b)

consider the motion along the Y-direction

Voy = initial velocity = Vo Sin45 = 14.5 Sin45

Ymax = maximum height gained = ?

a = acceleration = - 9.8

Vfy = final velocity at the highest point = 0

using the equation

V2fy = V2oy + 2 a Ymax

02 = (14.5 Sin45)2 + 2 (- 9.8) Ymax

Ymax = 5.4 m

c)

t' = time taken to reach maximum height

using the equation

Vfy = Voy + a t'

0 = 14.5 Sin45 + (- 9.8) t'

t' = 1.05 sec

along the X-direction displacement is given as

X = Vox t' = 14.5 Cos45 (1.05) = 10.8 m

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