Question

A student project required the design of a projectile launcher to launch a cricket ball from the ground vertically upward as fast as possible. One particular test saw the ball launch with an initial velocity of 18.0 m/s. Determine the time (in seconds) it took for the ball to reach the teacher on its upward journey if she was standing on top of a 14.0 m high building.

Take gravitational acceleration to be 9.81 m/s2.

Answer 1

Answer on Question #51493, Physics, Mechanics | Kinematics | Dynamics

A student project required the design of a projectile launcher to launch a cricket ball from the ground vertically upward as fast as possible. One particular test saw the ball launch with an initial velocity of $18.0~\mathrm{m/s}$. Determine the time (in seconds) it took for the ball to reach the teacher on its upward journey if she was standing on top of a $14.0~\mathrm{m}$ high building.

Take gravitational acceleration to be $9.81 \, \text{m/s}^2$.

Solution:

Kinematics equation

$$y = y_0 + v_0 t - \frac{1}{2} g t^2$$

where $y_0 = 0$ and $y = 14.0 \, \text{m}$ is distance, $v_0 = 18.0 \, \text{m/s}$ is initial velocity.

Hence,

$$\begin{array}{l} y = v_0 t - \frac{1}{2} g t^2 \\ \frac{9.81}{2} t^2 - 18 t + 14 = 0 \\ 9.81 t^2 - 36 t + 28 = 0 \\ \end{array}$$ t_{1,2} = \frac{36 \pm \frac{36^2 - 4 \cdot 9.81 \cdot 28}{2 \cdot 9.81} = \frac{36 \pm 14.046}{19.62} $$t_1 = 1.119 \approx 1.12 \, \text{s}$$ $$t_2 = 2.55 \, \text{s}$$

we choose the smaller period of time (motion upward)

Answer: $t = 1.12 \, \text{s}$