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Question

Secure! https://session.masten ngphysics.com/myct/itemView?assignmentProblemlD-91027430 Alternative Exercise 3.118 7 of 7 Part A A tennis ball rolls off the edge of a tabletop of height 0.720 m above the floor and strikes the floor at a point a horizontal distance 1.70 m from the edge of the table. You can ignore air Find the time of flight. Take the free fall acceleration to be g- 9.80 m/s2 Submit Part B Find the magnitude of the initial velocity. Take the free fall acceleration to be g= 9.80 m/s2 mis Submit Part C Find the magnitude of the velocity of the ball just before it strikes the floor Take the free fall acceleration to be g= 9.80 m/s*

Explanation / Answer

Part A -

Height of the table top, h = 0.72 m

use the expression -

s = u*t + (1/2)*g*t^2

=> 0.72 = 0 + 0.5*9.8*t^2

=> t = sqrt[0.72 / (0.5*9.8)] = 0.38 s

so the time of flight = 0.38 s

Part B -

Initially the ball has no vertical component of velocity.

So, the initial velocity of the ball = 1.70 / 0.38 = 4.47 m/s

Part C -

Vertical velocity of the ball just before the striking to the horiontal, Vy = 9.8*0.38 = 3.72 m/s

So, final velocity of the ball at the striking point, V = sqrt[Vx^2 + Vy^2] = sqrt[4.47^2 + 3.72^2] = 5.81 m/s

Part D -

Direction, theta = tan^-1(3.72 / 4.47) = 39.8 deg with the positive direction of x - axis.

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