1) A body projected vertically upwards crosses points A and B separated by 28 m with 1/3rd and 1/4th of the velocity respectively. What is the maximum height of reached above the ground.
2) A ball is thrown vertically upwards with an initial velocity such that it can reach the maximum height of 15 m. If at the same distance a stone is dropped from a height of 15 m, then the ratio of the distances traveled by them when they cross each other.
3) A cannon fires a shell with a speed of 84 m/s. When the cannon is inclined at an angle of 45°, the horizontal distance covered is ______.
4) A ball of mass m is dropped from a height h1, and after striking the ground, it bounces back to a height h2(h2<h1). If ratio of KE while passing a point at height h=h2/2 in the two directions is 2:1 what is the ratio of h1 to h2.
3371) v2=u2+2as
here, v= 1/4v and u= 1/3 v and s=28m a=g= -10m/s2 [-ve as it is retardation.]
Find v2.
Again v2=u2+2as
Now is the second case, when we throw up the ball. Here, v2=0 and u2=the value of the previous v2.
Find s.
Doing all the calculations, ans most probably is:
576 m
3372) For the first case,
v2=u2+2as
Here, v=0; a=10 m/s2; s=15 m.
Find u.
u comes out to be 17.32
The objects meet each other after some particular time. Now, for the ball, let distance be S1. Since total distance= 15m, therefore for the stone, the distance after which they will meet is 15-S1 m.
We know, S= ut+ 12at2
Replace with values.
For the ball, S= S1; u=17.32; a=10.
For the stone, S=15-S1; u=0; a=10.
So the equations are:
S1= 17.32t -5t2
15-S1= 5t2
Add the equations.
t comes out to be 0.87
Therefore S1=10.75
So, 15-S1 is= 4.25
So, their ratio= 10.754.25= 2.53...
3373) cos 45° = basehypotenuse
=> 1√2=base84
So, base= 59.4
Therefore, ans= 59.4 m every second...