Question

a ball is projected up with velocity u.it reaches a point in its path at times t1 and t2 seconds from the time of projection.then (t1+t2)

Answer 1

Answer:

Explanation:

This problem is of 2-D vertical motion.

In such problem, time taken for any moving object during its upward motion is the same as it takes for downward motion.

Now coming to the solution, which equation should you choose to solve the problem is easy when you know what is to be sorted and which quantities are given.

As here you have been provided with some part of vertical distance of the entire motion, the time after this point for ball to reach maximum height and coming back is given. And as this motion is under influence of the gravity, hence acceleration due to gravity is taken as 9.8 or 10.

So lets break the problem to decide how to solve it.

As in the first case, ball is thrown above with some initial velocity (And kinetic energy) with which it reaches maximum height h=x+25 m. And again descend back with the accelerating velocity.

Hence the equation you should use that include height, acceleration due to gravity and initial velocity. which is

v2=u2+2gh  

The sign of acceleration is positive because while descent it is moving along the gravity. In case it goes upwards the acceleration is against the gravity and hence the equation would change to  v2=u2−2gh  which means that the velocity is decelerating or reducing. Thus you get to know the next equation.

w=v+gt  while descent or w =v−gt  while ascend.

At the maximum height, the ball stops moving further and the velocity becomes 0.

Thus using the second equation considering the point that after 25 m of height it moves to the highest point, stops there (w=0) and returns to the same height of 25 m, with the same velocity v. we get to know v at height 25 m. i.e.

v=gh=10×2=20 m/s.  

Substituting this value of V in the equation 1, we get (for ascending motion we get)

u=v2+2gh−−−−−−−√=400+2×10×25−−−−−−−−−−−−−−√=900−−−√=30 m/s  

Hence the ball was thrown with the velocity of 30 m/s upwards. its velocity decreases to 20 m/s at the height of 25 m from the ground and at height h it reduces to zero. Again increases to 20 m/s at height of 25 m and when it reaches to the ground it is 30 m/s.