As shown in the figure below, a small, solid, uniform ball is to be shot from po

Question

As shown in the figure below, a small, solid, uniform ball is to be shot from point P so that it rolls smoothly along a horizontal path, up along a ramp, and onto a plateau. Then it leaves the plateau horizontally to land on a game board, at a horizontal distance d from the right edge of the plateau. The vertical heights are h1 = 4.50 cm and h2 = 1.60 cm. With what speed must the ball be shot at point P for it to land at d = 5.00 cm?
m/s

As shown in the figure below, a small, solid, uniform ball is to be shot from point P so that it rolls smoothly along a horizontal path, up along a ramp, and onto a plateau. Then it leaves the plateau horizontally to land on a game board, at a horizontal distance d from the right edge of the plateau. The vertical heights are h1 = 4.50 cm and h2 = 1.60 cm. With what speed must the ball be shot at point P for it to land at d = 5.00 cm?

Explanation / Answer

let the speed of ball at the top point is v

let the t is the time taken to fall


h2 = 0.5*g*t^2

t = sqrt(2*h2/g)

= sqrt(2*0.016/9.8)

= 0.05714 m/s

v = d/t

= 0.05/0.05714

   = 0.875 m/s

let u is the speed of the ball at the lower surface


according to conservation of enrgy

0.5*m*u^2 + 0.5*I*w1^2 = m*g*h1 +0.5*m*v^2 + 0.5*I*w2^2

for sphere, I = (2/5)*m*r^2

0.5*m*u^2 + 0.5*(2/5)*m*r^2*w1^2 = m*g*h1 + 0.5*m*v^2 + 0.5*(2/5)*m*r^2*w2^2

here, r*w1 = u and r*w2*v

0.7*m*u^2 = m*g*h1 + 0.7*m*v^2

0.7*u^2 = g*h1 + 0.7*v^2

u = sqrt(v^2 + g*h1/0.7)

u = sqrt(0.875^2 + 9.8*0.045/0.7)

u = 1.1814 m/s

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