Question

A student drops a ball from the top of a tall building; the ball takes 2.8 s to reach the ground. (a) What was the ball’s speed just before hitting the ground? (b) What is the height of the building?

(a) speed before hitting the ground = Answer
m/s

(b) Height of building = Answer
m

Answer 1

Answer on Question 70436, Physics, Other

Question:

A student drops a ball from the top of a tall building. The ball takes $2.8\,s$ to reach the ground.

(a) What was the ball’s speed just before hitting the ground?

(b) What is the height of the building?

Solution:

(a) We can find the ball’s speed just before hitting the ground from the kinematic equation:

$$v = v_0 + g t,$$

here, $v$ is the ball’s speed just before hitting the ground, $v_0$ is the initial velocity of the ball (since initially the ball starts falling from rest it will be equal to zero), $g = 9.8\,m/s^2$ is the acceleration due to gravity (we take the downwards as the positive direction, so the acceleration due to gravity will be with sign plus) and $t$ is the time that needs the ball to reach the ground.

Then, we get:

$$v = g t = 9.8\, \frac{m}{s^2} \cdot 2.8\, s = 27.44\, \frac{m}{s}.$$

(b) We can find the height of the building from another kinematic equation:

$$h = v_0 t + \frac{1}{2} g t^2,$$ $$h = \frac{1}{2} g t^2 = \frac{1}{2} \cdot 9.8\, \frac{m}{s^2} \cdot (2.8\, s)^2 = 38.42\, m.$$

Answer:

(a) $v = 27.44\, \frac{m}{s}$.

(b) $h = 38.42\, m$.

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