Question

A baseball is thrown horizontally off a very tall building with an initial velocity of 17.5 m/s find the balls horizontal and vertical positions and velocities after 2.4 seconds have passed? Thank you!

Answer 1

Answer:

x = 42 m

y = -28 m

vx = 17.5 m/s

vy = -23.5 m/s

Explanation:

if the throw point is the origin and UP and AWAY are the positive directions.

s = s₀ + v₀t + ½at²

v = u + at

horizontal position = 0 + 17.5 m/s(2.4 s) + ½(0)(2.4)² = 42 m

vertical position = 0 + 0(2.4) + ½(-9.8 m/s²)(2.4 s)² = -28 m

horizontal velocity  = 17.5 + 0(2.4) = 17.5 m/s

vertical velocity = 0 + (-9.8)(2.4) = -23.5 m/s