A boy drops a ball from the top of a building. Another boy standing at the ground calculates the time in which the ball hits the ground to be 4 seconds. If the ball accelerates uniformly at the rate of 10 ms–2 , find the velocity with which the ball will hit the ground. Also find the height of the building.
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Answers
Kinematic Equations from Integral Calculus
Let’s begin with a particle with an acceleration a(t) is a known function of time. Since the time derivative of the velocity function is acceleration,
\[\frac{d}{dt}v(t)=a(t),\]
we can take the indefinite integral of both sides, finding
\[\int \frac{d}{dt}v(t)dt=\int a(t)dt+{C}_{1},\]
where C1 is a constant of integration. Since
\[\int \frac{d}{dt}v(t)dt=v(t)\]
, the velocity is given by
\[v(t)=\int a(t)dt+{C}_{1}.\]
Similarly, the time derivative of the position function is the velocity function,
\[\frac{d}{dt}x(t)=v(t).\]
Thus, we can use the same mathematical manipulations we just used and find
\[x(t)=\int v(t)dt+{C}_{2},\]
where C2 is a second constant of integration.
We can derive the kinematic equations for a constant acceleration using these integrals. With a(t) = a a constant, and doing the integration in (Figure), we find
\[v(t)=\int adt+{C}_{1}=at+{C}_{1}.\]
If the initial velocity is v(0) = v0, then
\[{v}_{0}=0+{C}_{1}.\]
Then, C1 = v0 and
\[v(t)={v}_{0}+at,\]
which is (Equation). Substituting this expression into (Figure) gives
\[x(t)=\int ({v}_{0}+at)dt+{C}_{2}.\]
Doing the integration, we find
\[x(t)={v}_{0}t+\frac{1}{2}a{t}^{2}+{C}_{2}.\]
If x(0) = x0, we have
\[{x}_{0}=0+0+{C}_{2};\]
so, C2 = x0. Substituting back into the equation for x(t), we finally have
\[x(t)={x}_{0}+{v}_{0}t+\frac{1}{2}a{t}^{2},\]
which is (Equation).
Answer:
Given :-
Time taken for ball to reach the ground = 4 secs
Acceleration due to gravity = 10 m/s^2
Intial Velocity = 0 m/s ( since the ball starts from rest)
v = u + gt
v = 0 + 10 × 4
v = 40 m/s
h = ut +1/2gt^2
h = 0×4 + 1/2×10×4^2
h = 0 + 80
h = 80 m
Therefore, the velocity with which the ball hits the ground = 40 m/s and the height of the building = 80 m