A ball is thrown from ground level vertically upwards with an initial velocity of vi. Draw a velocity-time graph of the ball’s motion. Consider collisions with the ground are absolutely elastic and neglect air resistance.
One stone is thrown upwards from a point on the ground with an initial velocity v1 and simultaneously another stone is dropped from a height. They reach the ground at the same instant. Compare the velocities which they have attained.
A cannon ball is fired vertically upward with an initial speed of Vo. At the highest point it explodes into two pieces of equal masses. One of the pieces hits the ground with a speed of 2Vo.
What is the speed of the second piece when it hits the ground?
How long was the second piece in the air?
A body is thrown vertically upwards from the top of the tower. It reaches the ground in t1 seconds. If it is thrown vertically downwards from the same point with the same speed, it reaches the ground in t2 sec. Please prove If it is allowed to fall freely from the top of the tower, then the time it takes to reach the ground t is given by formula t=√(t1*t2).
Formula. Yf=Yi+Vi*t+a*t^2/2; Yf – final position; Yi – initial position; a=-g=-9.81 m/s^2; Yi=h – height of the tower; Vi – initial velocity
The body thrown vertically upwards .
The body thrown vertically downwards .
The body fall freely.
0=h-g*t^2/2 ; So, h=g*t^2/2.
Let’s use it for the first and second equations.
From the second equation
Use it for the first equation.
0=g*t^2/2+(g*t^2-g*t2^2)/(2*t2)*t1–g*t1^2/2 ; Divide by g
0=t^2/2+(t^2-t2^2)/(2*t2)*t1-t1^2/2; Multiply by 2
0=t^2+(t^2-t2^2)/t2*t1-t1^2; Multiply by t2
0=t^2*(t1+t2)-t1*t2*(t1+t2) ; Divide by (t1+t2)