calculus

A stone is dropped from the upper observation deck of a tower, 450 m above the ground. (Assume *g* = 9.8 m/s^2)

(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.)

(d) If the stone is thrown downward with a speed of 5 m/s, how long does it take to reach the ground? (Round your answer to two decimal places.)

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Since the question is based on calculus we need to know that

dS/dt = V

S= distance in meters

t= time in seconds

V= velocity in m/s

In general terms S= U*t+0.5*a*t^2

U = initial velosity

a= accelaration

a)

So if you put S= U*t+0.5*a*t^2 ------------- (1) when the stone releases from tower;

Since the stone falls in free fall U=0 and a=g

Then S=0.5*g*t^2 --------------- (2)

The distance (in meters) of the stone above ground level will be

450-S

**Therefore distance (in meters) of the stone above ground level at the time t; S1= 450-0.5*g*t^2**

b) Distance (in meters) of the stone above ground level at the time t; S1= 450-0.5*g*t^2

When the stone reaches ground S1=0

Then 450-0.5*g*t^2 = 0

0.5*g*t^2 = 450

t^2 = 450/(0.5g)

t^2 = 91.837

t= sqrt(91.837)

t= +9.58 OR -9.58

Since time is positive** t= 9.58s**

c)

If the velocity that the stone hits the round is V

dS/dt = V as stated earlier.

V =dS/dt

=d(0.5*g*t^2)/dt

= 0.5*g*2t

At t=9.58s stone falls to the ground.

Then **V= 0.5*9.8*9.58 = 46.94 m/s**

d)

If the stone thows at 5m/s; then U= 5m/s as in (1)

S= U*t+0.5*a*t^2

S= 450m (tower height)

a = g (accelaration due to gravity)

Then;

450 = 5*t+0.5*9.8*t^2

0= 0.5*9.8*t^2+5*t-450

So t1= (-5+sqrt(5^2-4*0.5*9.8*-450)/(2*0.5*9.58)

t2 = (-5-sqrt(5^2-4*0.5*9.8*-450)/(2*0.5*9.58)

t1= 9.09

t2= -10.11

**since the time is positive the time taken by the stone to reach ground is 9.09s**

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