00:01
The solution we are given that a rock is thrown upward from a level ground in such a way that the maximum height of its height is equal to the horizontal range.
00:08
For that, we know the formula for the maximum height.
00:13
It is v0 square, sine square theta over 2g where v .0 is the initial velocity and teta is the angle at which it is thrown, which is something which we have to find.
00:23
This is about project in motion.
00:25
And the maximum range is given as v0 square, sine 2 theta.
00:31
Over g so if they are equal so let's equate them this means that h is equal to g which means that v0 squared sine square theta over 2g is equal to v0 square sine 2 theta over g so this g is cancelled v0 square is cancelled what i will left with is sine square theta over 2 and sine 2 theta can be written as 2 sine teta cost theta all right simplify this further, in fact, one of the signs gets cancelled and we are just left with one sign.
01:09
So one sign that is a sine theta over 2 is equal to 2 cost theta, which means that if we do a cross multiplication, sine over cost is tan, so 10theta is equal to 4, which means that the value of theta is, grab my calculator here, tan inverse 4, which is 75 .96 degrees.
01:29
So this is the required angle.
01:31
Part b asks that what if this situation was on a different planet? so a different planet means there's a different value of g.
01:40
But this solution is independent of the value of g because if you notice carefully, this g automatically is getting cancelled over here...