Physics

Problem

A ball is launched vertically upward from ground level (

x = 0

) with an initial speed of 20 m/s at

t = 0

. For the first 1.5 s the only acceleration acting on the ball is gravity, directed downward with magnitude

g = 9.8\,\text{m/s}^2

. At

t = 1.5\,\text{s}

, small upward thrusters embedded in the ball ignite, producing an additional constant upward acceleration of 4.0 m/s

^2

for the next 0.80 s. Immediately after the thrusters shut off (

t = 2.30\,\text{s}

), the ball continues to move only under the influence of gravity until it returns to the launch point. Neglect air resistance. (a) Determine the ball’s velocity just before the thrusters ignite at

t = 1.5\,\text{s}

. (b) Determine the ball’s velocity immediately after the thrusters shut off at

t = 2.30\,\text{s}

. (c) Calculate the additional time required for the ball to reach its maximum height after

t = 2.30\,\text{s}

. (d) Determine the maximum height of the ball above the launch point. (e) Calculate the total time interval from launch until the ball returns to ground level.

Intuitive explanation

This problem asks you to track motion through three distinct phases, each with different accelerations—the key is treating each segment separately using kinematic equations, then using the final state of one phase as the initial condition for the next. First, find where the ball is and how fast it's moving when gravity alone acts for 1.5 s, then apply the net upward acceleration (thruster plus gravity) for the next 0.80 s to get new position and velocity, and finally use those values to determine the time it takes gravity alone to bring the ball back to ground level.

Intuitive explanation

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