00:01
Hello, in this question we are given that a projectile is launched from the ground and at a constant launch angle.
00:10
If the speed with which the projector was launched is changed, what will be effect on the range and the maximum height.
00:18
Okay, so first let us evaluate the expression that will be given by any general velocity v, that is launched at an angle of theta.
00:27
So we can calculate the value of maximum height and range.
00:32
This was the maximum height h and this was the range r so for maximum height h what you will do is we will take the vertical component of the speed okay that would be what v sine theta okay initially in vertical direction initial velocity that may solve for the vertical okay initial velocity is v sine theta that is equal to u and acceleration is minus g that is equal to a, okay, because acceleration is always acting vertically downward.
01:10
So, we need to calculate the height at the highest point.
01:14
At highest point, velocity would be, final velocity would be zero.
01:18
So, v square minus u square is equal to 2a .s.
01:23
This would be used to calculate the peak height...