#14 /54.8° 66.3° A ball is launched off a 4-m high cliff with a speed of 14 m/s

Question

#14 /54.8° 66.3° A ball is launched off a 4-m high cliff with a speed of 14 m/s at an angle of 35. Which of the following statements is true about the ball's velocity at the peak of the trajectory? a) vx = 0 m/s, Vy = 0 m/s Vs= 11.5 m/s, Vy = 8.0 m/s c) 'v,-11.5 m/s, Vy = 0 m/s d) Vx= 8.0 m/s, Vy = 0 m/s e) vx 0 m/s, vy-9.8 m/s A ball is launched off a 4-m high cliff with a speed of 14 m/s at an angle of 35. What is the ball's velocity just before landing? a) Vx = 11.5 m/s, v,-8.0 m/s -11.5 m/s, v,--8.0 m/s c) v,--11.5-m/s, vy = 8.0 m/s d) Vx 11.5 m/s, vy 0m/s e) vx = 0 m/s, Vy = 0 m/s

Explanation / Answer

while the ball was going up,the ball was acted upon by only gravitational force,so accelaration due to gravity is only in vertical direction

there is no accelaration along horizontal direction.i.e the horizontal component of speed of the ball is constant through out the journey

at the peak of the trajectory the verical component of velocity becomes zero

Horizontal component of velocity is Vx = 14*cos(35) = 11.5 m/s

and vy = 0 m/sec

hence the answer is C) Vx = 11.5 m/s,Vy = 0 m/s

---------------------------------------------------------

vx = 14*cos(35) = 11.5 m/sec

time of flight is T = 2*Vo*sin(35)/g = (2*14*sin(35))/9.8 = 1.64 sec

using

Vy = Voy - (g*T)

Vy = (14*sin(35))-(9.8*1.64) = -8 m/sec

so the answer is

Vx = -11.5 m/sec and Vy = -8 m/sec

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