Question

A stone is dropped from the upper observation deck of a tower, 600 m above the ground. (Assume g = 9.8 m/s2.) (a) Find the distance (in meters) of the stone above ground level at time t. (b) How long does it take the stone to reach the ground? (Round your answer to two decimal places.) (c) With what velocity does it strike the ground? (Round your answer to one decimal place.) Remember that velocity requires direction. (d) If the stone is thrown downward with a speed of 7 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.)

Answer 1

Answer:

a) $D = 600 -4.9t^2$

b) 11.06 seconds

c) 108.39 m/s

d) 10.37 m/s

Step-by-step explanation:

Given:

Distance, s = 600 m

Acceleration, a = g = 9.8 $m/s^2$

a) Distance of stone above ground level at time 't'.

First of all, we need to find the distance traveled in time 't' and then we will subtract it from 600 to find the answer.

The formula is given as:

$s=ut+\dfrac{1}{2}at^2$

where u is the initial velocity which is 0 in this case.

$s=0\times t+\dfrac{1}{2}\times 9.8 \times t^2\\s =4.9t^2$

Distance of stone above ground level at time 't',

$D = 600 -4.9t^2$

b) Time taken by stone to reach the ground. i.e. D = 0

Using above equation, putting D = 0

$0 = 600 -4.9t^2\\\Righttarow 4.9t^2 = 600\\\Rightarrow t = \sqrt{\dfrac{6000}{49}} = 11.06\ sec$

c) Velocity with which it strikes the ground i.e. $v=?$

Using the formula:

$v=u+at$

$v = 0 +9.8 \times 11.06\\v = 108.39\ m/s$

d) If initial velocity, u = 7 m/s, time taken to reach the ground = ?

In this case total distance traveled = 600 m

$s=ut+\dfrac{1}{2}at^2$

$600=7 t+\dfrac{1}{2}\times 9.8t^2\\\Rightarrow 600=7 t+4.9t^2\\\Rightarrow 4.9t^2+7 t-600=0\\\Rightarrow 49t^2+70 t-6000=0$

Solving the above equation:

t = 10.37 seconds

The answers are:

a) $D = 600 -4.9t^2$

b) 11.06 seconds

c) 108.39 m/s

d) 10.37 m/s