Question

Find the velocity of a projectle at the highest point, if it is projected with a speed 15 ms
in the direction 45° above horizontal

Ans is 15/root2

PLS GIVE CORRECT EXPLANATION ​

Answer 1

At highest point of projectile motion the velocity of the object is zero as the vertical component of velocity momentarily is zero so it moves due to horizontal component of velocity only i.e.

vx=vcosθ.

v(horizontal) = 15cos45°= 15/√2

Answer 2

Answer: 15/√2

Explanation

Initial velocity is given that is u= 15m/s

At the highest point only horizontal velocity is acting on the body which is given by Ucos(theta)

So as theta is 45° Ucos = 15×cos45°= 15/√2