2D Kinematics - Calculating Speed Question
A volleyball is sent over a net, having been launched at a speed of 4.2 m/s at an angle 44.2 degrees above the horizontal, from the ground. What is the ball’s speed, in m/s, 2.42 seconds after being launched?
Homework Answers
Formula: V = √[(v0cosθ)2 + (v0sinθ - gt)2]
Here, V = the terminal speed
v0 = the initial speed = 4.2 m/s
θ = angle of launching = 44.2ο
t = time = 2.42 sec.
∴ V = √[(4.2*cos(44.2))2 + (4.2*sin(44.2) - 9.8*2.42)2]
= √[(3.01)2 + (2.93 - 23.72)2]
= 21.005 m/s
the ball's speed is 21.005 m/s at 2.42 seconds after being launched.
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