Give the expression for the time of flight and the maximum height reached by the projective.

Deduce an expression for the time of flight, vertical height and range in case projectile motion.

A stone is projected in air. Its time of flight is 3s and range is 150m. Maximum height reached by the stone is (Take, g = 10 ms^(-2))

A stone is projected in air. Its time of flight is 3s and range is 150m. Maximum height reached by the stone is (Take, g = 10 ms^(-2))

What is a projectile ? Obtain an expression for: maximum height , time of flight and the horizontal range when a projectile is fired at an angle theta with the horizontal.

Two bodies are projected at angles 30^(@) and 60^(@) to the horizontal from the ground uch that the maximum heights reached by them are equal. Then (a) Their times of flight are equal (b) Their horizontal ranges are equal (c) The ratio of their initial speeds of projection is sqrt(3) :1 (d) Both take same time to reach the maximum height. Mark the answer is

If T be the total time of flight and H be the maximum height attained by a stone from the point of projection, then H//T will be ( u=projection velocity and theta =projection angle )

Two bodies are projected at angles 30^(@) and 60^(@) to the horizontal from the ground such that the maximum heights reached by them are equal then a) Their times of flight are equal b) Their horizontal ranges are equal c) The ratio of their initial speeds of projection is sqrt3:1 d) Both take same time to reach the maximum height.

Introduction || Time OF flight, Range, Maximum Height