Question

A projectile is projected from ground with velocity 20m/s at angle of 45^(@) with horizontal.Height of the projectile above the ground when its speed is minimum is g=10m/s^(2)​

Answer 1

Answer:

20

Explanation:

As we know, that velocity is minimum i.e 0 at " maximum height"

So basically we have to find maximum height of projectile from ground

Let maximum height " hmax"

for calculation hmax , apply formula

hmax = u²/2g , where u= initial velocity

so from here hmax = 20