Try this problem, useful for Physics Olympiad based on Vertical Motion.
The Problem: Vertical Motion
A boy is standing on top of a tower of height 85m and throws a ball in the vertically upward direction at a certain speed. If 5.25 secs later he hears the ball hitting the ground, then the speed with which the boy threw the ball is ( take g=10m/s2, speed of sound in air=340m/s)
(A) 6m/s
(B) 8m/s
(C) 10m/s
(D) 12m/s
Solution:
Time taken by sound= (\frac{85}{340})=0.25secs
Time taken by the ball= 5.25-0.25=5s
Now,
Let us assume that the ball is thrown upwards with a velocity u. We know, (s=ut+\frac{1}{2}gt^2) where s is the distance covered,u is the initial velocity, g is the acceleration due to gravity and t is the time taken. Here, time taken t=5. Since the ball is thrown upwards, we have $$85=-5u+\frac{1}{2}\times10\times25$$
Hence, u is 8m/s.
Position of a Particle
Try this problem useful for the Physics Olympiad based on the Position of a Particle.
The problem:
A particle of mass m is subject to a force $$ F(t)=me^{-bt}$$. The initial position and speed are zero. Find (x(t)).
Solution: In the given problem $$ \ddot{x}=e^{-bt}$$
Integrating this with respect to time gives $$ v(t)=-\frac{e^{-bt}}{b}+A $$ ( A is the constant of integration)
We integrate again with respect to x.
$$ x(t)=\frac{e^{-bt}}{b^2}+At+B$$ ( B is the constant of integration)
The initial condition ( v(0)=0),gives (\frac{-1}{b}+A=0) $$\Rightarrow A= \frac{1}{b}$$
The intial condition $$ x(0)=0$$, gives $$\frac{1}{b^2}+B=0$$ $$\Rightarrow B=-\frac{1}{b^2}$$
