Question

2. Velocity of a projectile is 10m/s. At what angle to the horizontal should it be projected so that it covers maximum horizontal distance?

Answer 1

Answer:

Explanation:

Hi!

The formula for the  max range is{ Q is the angle}

u^2sin2Q /g

now for max range, the angle becomes 45 (Sin2Q=90)(sin 90=1)

so the formula obtained for max r is

R=u^2/g

On substituting value

g=10m/s^2

R=100/10

R is 10 m

Answer 2

HI DEAR SHRESTA HERE !!!!!!PLS FOLLOW ME FRND....

The formula for the  max range is{ Q is the angle}

u^2sin2Q /g

now for max range, the angle becomes 45 (Sin2Q=90)(sin 90=1)

so the formula obtained for max r is

R=u^2/g

On substituting value

g=10m/s^2

R=100/10

R is 10 m