Question

A ball is thrown vertically upwards from the top of the tower with a speed of 40m/s returns back to ground level in 10s.The height of the tower is

Answer 1

Answer on Question 63101, Physics, Mechanics

Question:

A ball is thrown vertically upwards from the top of the tower with a speed of $40\ \mathrm{m/s}$ returns back to ground level in $10\ \mathrm{s}$. What is the height of the tower?

Solution:

Let's take the upwards as the positive direction. Then, we can find the height of the tower from the kinematic equation:

$$h = v_0 t + \frac{1}{2} g t^2,$$

here, $h$ is the height of the tower, $v_0$ is the initial velocity of the ball, $g = -9.8\ \mathrm{m/s^2}$ is the acceleration due to gravity, $t$ is the time of flight.

Finally, we can calculate the height of the tower:

$$h = v_0 t + \frac{1}{2} g t^2 = 40\ \frac{\mathrm{m}}{\mathrm{s}} \cdot 10\ \mathrm{s} + \frac{1}{2} \cdot \left(-9.8\ \frac{\mathrm{m}}{\mathrm{s^2}}\right) \cdot (10\ \mathrm{s})^2 = -90\ \mathrm{m}.$$

The sign minus appears here because we take the upwards as the positive direction. So, the height of the tower is $90\ \mathrm{m}$.

Answer:

$$h = 90\ \mathrm{m}.$$

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