Question

You throw a ball upward with an initial speed of 3.0 m/s from an initial height of 1.5 m. After you throw the ball, its acceleration is 9.81 m/s2 downward.

Taking upward to be the positive direction, write the position-time equation for the ball's motion.

Express your answer in terms of t. t is expressed in seconds.

Answer 1

Here,

for the ball

initial speed ,u = 2 m/s

a = -9.81 m/s^2

y0 = 1.5 m

Using second equation of motion

y = y0 + u * t + 0.50 * a * t^2

y = 1.5 + 3t - 0.50 * 9.8 * t^2

y = 1.5 + 3t - 4.9t^2

the expression is 1.5 + 3t - 4.9t^2