(iv) Horizontal Range (R):l an the horizontal dis

1.	(iv) Horizontal Range (R):l an the horizontal dis
Answer» The initial velocity and angle of projection decides the range as well as the height it reaches. Let V be the velocity and A be the angle, then the velocity is split into horizontal and vertical components as VcosA and VsinA. This problem is solved easily if we understand this. That the vertical component decreases due to gravity and reaches zero at the top, then reverses and reaches its initial value when the projectile hits ground. And that the horizontal component remains constant neglecting wind. So the range will be time taken for the projectile to go to the top and reach ground again Now we have to use velocity acceleration formula v=u+at If t is the time taken from start to reach the top and g is the gravity constant then 0=VsinA-gt t=VsinA/g Total time = 2 VsinA/g So range= 2 VsinA/g. VcosA you have nice thinking capability

Answer»

The initial velocity and angle of projection decides the range as well as the height it reaches.

Let V be the velocity and A be the angle, then the velocity is split into horizontal and vertical components as VcosA and VsinA.

This problem is solved easily if we understand this.

That the vertical component decreases due to gravity and reaches zero at the top, then reverses and reaches its initial value when the projectile hits ground.

And that the horizontal component remains constant neglecting wind.

So the range will be

time taken for the projectile to go to the top and reach ground again

Now we have to use velocity acceleration formula

v=u+at

If t is the time taken from start to reach the top and g is the gravity constant

then

0=VsinA-gt

t=VsinA/g

Total time = 2 VsinA/g

So range= 2 VsinA/g. VcosA

you have nice thinking capability

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