A home run is hit in such a way that the baseball just clears a wall 16 0 m high

Question

A home run is hit in such a way that the baseball just clears a wall 16 0 m high, located 36 m from home plate. The ball is hit at an angle of 37.0 to the horizontal, end air resistance is negligible. (Assume that the ball is hit at a height of 1.0 m above the ground.) Find the initial speed of the ball. Find the time it takes the ball to reach the wall. Find the velocity components of the ball when it reaches the wan. x-component y-component Find the speed of the ball when it reaches the wall. A car is parked on a steep incline, making an angle of 37.0 below the horizontal and over.ook.ng the ocean, when it. brake, fail and 4 begins to roll Starting from rest at t 0, the car rolls down the incline with a constant accelerate of 3.77 m/s^2, traveling to the edge of a vertical diff The Cliff is 30.0 m above the ocean. Find the speed of the car when it reaches the edge of the cliff

Explanation / Answer

h = 16 m ; d = 136 m ; theta = 37 deg ; h' = 1 m

a)the time required to reach the distance wall at d = 136 m would be:

t = d/Vix cos(theta)

t = 136/Vi cos37 = 170.3/Vi

we know, s= ut + 1/2 at^2

s = 16 - 1 = 15 ; a = -g ; t = 170.3/Vix

15 = (Vi sin37) (170.3/Vi) - 0.5 x 9.8 x (170.3/Vi)^2

15 = 102.5 - 142110.2/Vi^2

(102.5 - 15)Vi^2 = 142110.2

Vi = 40.30 m/s

Hence, Vi = 40.3 m/s

b)t = 170.3/Vi

t = 170.3/40.3 = 4.23 s

Hence, t = 4.23 s

c)Vix = Vi cos(theta)

Vix = 40.3 x cos37 = 32.2 m/s

Viy = 40.3 x sin37 = 24.3 m/s

Vix = 32.2 m/s ; Viy = 24.3 m/s

d)from eqn of motion:

v = u + at

Vy = Viy + (-g)t

Vy = 24.3 - 9.8 x 4.23 = -17.2

V = sqrt [(32.2)^2 + (-17.2)^2] = 36.5 m/s

Hence, V = 36.5 m/s

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