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

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)`.)

Step-by-step 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} = (1)/(2)* 9.81\; {\rm m\cdot s^(-2)} * t^(2) + 0\; {\rm m \cdot s^(-1)} * t`.

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

`\displaystyle t^(2) = (20.0\; \rm m)/((1/2) * 9.81\; \rm m\cdot s^(-2))`.

Since
`t > 0`:

`\begin{aligned}t &= \sqrt{(20.0\; \rm m)/((1/2) * 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) * 2.01928\; \rm s \\ &\approx 19.8\; \rm m \cdot s^(-1) \end{aligned}`.

Answer 2

Hi there!

We can begin by using the derived kinematic equation:

`t = \sqrt{(2h)/(g)`

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

`t = \sqrt{(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}`