A ball is thrown horizontally from the top of a 69.0-m building and lands 119.8

Question

A ball is thrown horizontally from the top of a 69.0-m building and lands 119.8 m from the base of the building. Ignore air resistance. (Assume the ball is thrown in the +x direction and upward to be in the +y direction.) (a) How long (in s) is the ball in the air? (b) What must have been the initial horizontal component of the velocity (in m/s)? (Indicate the direction with the sign of your answer.) m/s (c) What is the vertical component of the velocity (in m/s) just before the ball hits the ground? (Indicate the direction with the sign of your answer.) (d) What is the velocity (in m/s) (including both the horizontal and vertical components) of the ball just before it hits the ground? magnitude direction counterclockwise from the +x-axis

Explanation / Answer

(a) height of the building, h = 69.0 m

suppose the ball remains in air for t sec.

use the expression -

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

put the values -

69 = 0 + 0.5*9.81*t^2

=> t = 3.75 s

(b) here, d = 119.8 m

so, the initial horizontal component of the velocity = 119.8 / 3.75 = 31.95 m/s

(c) vertical component of the velocity just before the ball hits the ground -

use the expression -

v = u + a*t

v = 0 + 9.81*3.75 = 36.79 m/s

(d) Final velocity of the ball just before it hits the ground -

v = sqrt[31.95^2 + 36.79^2] = 48.73 m/s

direction, theta = tan^-1(36.79 / 31.95) = 49.0 deg counterclockwise from the +x - axis.

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