A tower is 24 m high. A stone is thrown from the tower with an initial velocity of 24.5 m/s. Calculate how long it would take the stone to reach the ground at the foot of the tower
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A tower is 24 m high. A stone is thrown from the tower with an initial velocity of 24.5 m/s. Calculate how long it would take the stone to reach the ground at the foot of the tower

You need to specify the direction in which the stone is thrown.

Is it thrown vertically upwards, vertically downwards, or horizontally from the tower?

Assume it is thrown vertically upwards.

The equation to use is,

s = ut + (½).at^2

Take downwards direction as positive. Then

s = 24m

u = -24.5 m/s

a = 9.81 m/s^2

t = ? secs

Then,

24 = -24.5t + (1/2)9.81t^2

9.81t^2 – 49t – 48 = 0

Using the quadratic formula

t = (49 ± √((-49)^2 – 4*9.81*(-48)))/(2*9.81)

t = (49 ± √(2401 + 1883.52)/(19.62)

t = (49 ± √(4284.52)/(19.62)

t = (49 ± 65.456)/(19.62)

Ignoring negative results

t = 5.83365

Time to reach the ground is 5.83 secs

If the stone is thrown vertically downwards, then you will need to use u = 24.5 in the given formula.

If the stone is projected horizontally from the tower, then you would to use u = 0 in the given formula.

 

by Level 11 User (81.5k points)

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