Question

A stone is projected vertically upwards from the top of a tower 73.5m high with a velocity of 24.5 m/s. Find the time taken by the stone to reach the foot of the tower. ​

Answer 1

t=?

a= -9.8m/s

u= 24.5m/s

s= -73.5m

s=ut+1/2at^2

-73.5=24.5t+(1/2-9.8t^2)

-73.5=24.5t-(4.9t^2)

(4.9t^2)-24.5t-73.5=0

Dividing all throughout by 4.9

(t^2)-5t-15=0

Then we converted it to a second degree equation

Solve this using

Answer 2

Answer:

The time taken by the stone to reach the foot of the tower is 7.1sec

Explanation:

Given that,

A stone is projected vertically upwards from the top of a tower 73.5m high with a velocity of 24.5 m/s.We are required to find the time taken by the stone to reach the bottom of the tower.To solve this we shall use the equation of motion as shown

$s = ut + \frac{1}{2} a {t}^{2}$

In this equation for our problem,

s=-73.5,u=24.5,a=-g

Substituting these values in the above relation,we get

$- 73.5 = 24.5t + \frac{1}{2} ( - 9.8) {t}^{2} \\ = > - 73.5 = 24.5t - 4.9 {t}^{2} \\ = > 4.9 {t}^{2} - 24.5t - 73.5 = 0 \\ = > {t}^{2} - 5t - 15 = 0$

Therefore the solution of the above equation is

$t = \frac{5± \sqrt{ {5}^{2} - 4(1)( - 15) } }{2(1)} = \frac{5± \sqrt{85} }{2}$

As time cant be negative hence the value of t is

$t = \frac{5 + \sqrt{85} }{2} = \frac{5 + 9.21}{2} = 7.1 \: sec$

Therefore,The time taken by the stone to reach the foot of the tower is 7.1sec

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