Anonymous
Anonymous asked in Science & MathematicsPhysics · 4 weeks ago

The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 12 m/s at an angle 47 degreesground.?

The ball leaves Julie's hand a distance 1.5 meters above the ground with an initial speed of 12 m/s at an angle 47 degrees; with respect to the horizontal. Sarah catches the ball 1.5 meters above the ground.

1)What is the horizontal component of the ball’s velocity right before Sarah catches it?

2)What is the vertical component of the ball’s velocity right before Sarah catches it? 

3)What is the time the ball is in the air? 

4)What is the distance between the two girls? 

5)After catching the ball, Sarah throws it back to Julie. However, Sarah throws it too hard so it is over Julie's head when it reaches Julie's horizontal position. Assume the ball leaves Sarah's hand a distance 1.5 meters above the ground, reaches a maximum height of 8 m above the ground, and takes 1.464 s to get directly over Julie's head.What is the speed of the ball when it leaves Sarah's hand?

6)How high above the ground will the ball be when it gets to Julie?

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  • 4 weeks ago
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    ) As the horizontal component of velocity is free from gravitational force => It remain constant during the flight of the projectile

    =>Ux = ucosθ = 12 x cos47* = 8.18 m/s

    2) As the horizontal height is same at the two ends of projectile=>the vertical velocity will be same at both the ends

    =>Vy = Uy = usinθ = 12 x sin47* = 8.78 m/s

    3) As the horizontal height is same at the two ends of projectile, the total time of flight

    =>by T = 2usinθ/g

    =>T = [2 x 12 x sin47*]/9.8 = 1.79 sec

    4) By R = u^2sin2θ/g = [(12)^2 x sin(2 x 47)*]/9.8 = 14.66 m

    5) Let the initial velocity of the ball is u m/s and the angle of projection is θ, let the time taken to cover 14.66m is t sec

    =>By R = (Ux) x t

    =>14.66 = ucosθ x 1.464

    =>ucosθ = 10.01 -------------------------(i)

    The maximum height attained by the ball is 8 m thus the maximum height of the projectile is (8 - 1.5m) = 6.5m

    =>By H = (usinθ)^2/2g

    =>usinθ = √2gH = √[2 x 9.8 x 6.5] = 11.29 -----------------(ii)

    By (i)^2 + (ii)^2:

    =>u = √[(10.01)^2 + (11.29)^2] = 15.09 m/s

    6) Let the ball gains h meter in 1.464 sec

    =>By s = ut - 1/2gt^2

    =>h = 11.29 x 1.464 - 1/2 x 9.8 x (1.464)^2 = 6.03 m

    Thus the height grom the ground(H) = 1.5 + h = 1.5 + 6.03 = 7.53 m

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