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...
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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