Question

. A ball thrown vertically upwards with a speed of 19.6 ms-1 from the top of a tower returns to the earth in 6 seconds. Find the height of the tower. (g = 9.8 ms-2)

Answer 1

Answer:

Explanation:

Hopeit help u

Answer 2

Answer:

The height of the tower is 59m.

Explanation:

We will use the first and second equations of motion to solve this question.

The first equation of motion

$v=u+at$        (1)

The second equation of motion

$s=ut+\frac{1}{2}at^2$      (2)

v=initial velocity

u=final velocity

a=acceleration

t=time

s=height

From the question we have,

u=19.6 m/s

v=0

t=6seconds

First, let's find the acceleration of the ball using equation (1),

$0=19.6+a\times 6$

$a=\frac{-19.6}{6}=-3.26m/s^2$          (3)

From equation (2) we will get the height of the tower i.e.

$s=19.6\times 6-\frac{1}{2}\times 3.26\times 6^2=58.9m\approx59m$

Hence, the height of the tower is 59m.