Question

A ball is dropped from a 20.0 m high tower.

a. How long will it take the ball to reach the ground?

b. What is the velocity of the ball just before it hits the ground?

Answer 1

Hi there!

We can begin by using the derived kinematic equation:

$t = \sqrt{\frac{2h}{g}$

Plug in the given values and let g = 9.8 m/s²:

$t = \sqrt{\frac{2(20)}{g}} = \boxed{2.02 s}$

Now, we can solve for its final velocity using the equation:

$v_f = v_i + at$

It is dropped from rest, so vi = 0 m/s.

$v_f = at$

$v_f = 9.8(2.02) = \boxed{19.796 m/s}$

Answer 2

Answer:

a. The ball would reach the ground in approximately $2.02\; \rm s$.

b. The velocity of the ball right before landing would be approximately $19.8\; \rm m\cdot s^{-1}$.

(Assumptions: the ball was dropped with no initial velocity; air resistance on the ball is negligible; $g = 9.81\; \rm m\cdot s^{-2}$.)

Explanation:

Under these assumptions, the acceleration of this ball would be constantly $a = g = 9.81\; \rm m\cdot s^{-2}$ (same as the gravitational field strength) during the descent.

Displacement of the ball: $x = 20.0\; \rm m$.

Initial velocity of the ball: $v_{0} = 0\; \rm m\cdot s^{-1}$.

Let $t$ denote the duration of this descent.

The SUVAT equation $x = (1/2)\, a\, t^{2} + v_{0}\, t$ relates the known quantities $a$, $x$, and $v_{0}$ to the unknown $t$.

Substitute the known quantities into this equation and solve to find the value of $t\!$:

$\displaystyle 20.0\; {\rm m} = \frac{1}{2}\times 9.81\; {\rm m\cdot s^{-2}} \times t^{2} + 0\; {\rm m \cdot s^{-1}} \times t$.

$\displaystyle 20.0\; {\rm m} = \frac{1}{2}\times 9.81\; {\rm m\cdot s^{-2}} \times t^{2}$.

$\displaystyle t^{2} = \frac{20.0\; \rm m}{(1/2) \times 9.81\; \rm m\cdot s^{-2}}$.

Since $t > 0$:

$\begin{aligned}t &= \sqrt{\frac{20.0\; \rm m}{(1/2) \times 9.81\; \rm m\cdot s^{-2}}} \\ &\approx 2.01928\; \rm s\\ &\approx 2.02\; \rm s \\ & (\text{Rounded to 2 sig. fig.})\end{aligned}$.

Since the acceleration of this ball is constant, the velocity of the ball right before landing would be:

$\begin{aligned}v_{1} &= a\, t \\ &\approx 9.81\; \rm m\cdot s^{-2} \times 2.01928\; \rm s \\ &\approx 19.8\; \rm m \cdot s^{-1} \end{aligned}$.