Help 2 3 456 78 2/2 1/2-/3-4-2-2-20/3 320(15.0%) Points Scoring 16 3.4 P 038 My

Question

Help 2 3 456 78 2/2 1/2-/3-4-2-2-20/3 320(15.0%) Points Scoring 16 3.4 P 038 My A football quarterback is moving straight backward at a speed of 1.00 m/s when he thraws a pass to a player 16.0 m straight downfield. (a) If the ball is thrown at an angle of 25.09 relative to the ground and is caught at the same height as it is released, what is its initial speed relative to the groundi m/s (b) How long does it take to get to the receiver? (c) What is its maximum height above its point of release? 0074 Reading

Explanation / Answer

a)

along the y-direction :

Voy = initial speed of ball along vertical direction = Vo Sin25

t = time of travel

Y = displacement = 0

a = acceleration = - 9.8

using the equation

Y = Voy t + (0.5) a t2

0 = (Vo Sin25 ) t + (0.5) (-9.8) t2

t = (Vo Sin25 )/4.9 eq-1

Vox = initial speed of ball along horizontal direction = Vo Cos25 - 1

X = distance travelled = 16 m

using the equation

X = Vox t

16 = (Vo Cos25 - 1) ((Vo Sin25 )/4.9)

Vo = 14.86 m/s

b)

using eq-1

t = (Vo Sin25 )/4.9 = (14.86 Sin25)/4.9 = 1.3 sec

c)

along the Y-direction :

Ymax = maximum height

Vfy = final velocity at the maximum height = 0

using the equation

Vfy2 = Voy2 + 2 a Ymax

02 = (14.86 Sin25)2 + 2 (-9.8) Ymax

Ymax = 2.01 m

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