Physics · Motion In A Plane · NEET
The vertical motion is symmetric. The time to go up to the highest point is t_up = u sin(theta)/g, because at the top the vertical velocity becomes zero. The time to come back down is exactly the same, t_down = u sin(theta)/g. Total time of flight = t_up + t_down = 2u sin(theta)/g. The 2 simply adds the up-trip and the down-trip together.
No. Gravity acts only vertically, so only the vertical component u sin(theta) decides how long the body stays in the air. The horizontal component u cos(theta) moves the body sideways but never pulls it down or holds it up. Two balls dropped and thrown horizontally from the same height land at the same time.
No. Time to reach maximum height is only half the trip: t = u sin(theta)/g. Time of flight is the full up-and-down journey: T = 2u sin(theta)/g. So T = 2 times the time to reach the top. Students lose easy NEET marks by mixing these two up.
When theta = 0 (thrown horizontally from a cliff of height h), the formula 2u sin(theta)/g gives zero because sin 0 = 0. That formula is only for a projectile that lands at the same height it left. For a horizontal launch from height h you use h = (1/2) g t^2, so t = sqrt(2h/g).
Convert first. If the angle with the vertical is phi, then the angle with the horizontal is theta = 90 - phi. The vertical component becomes u cos(phi). So T = 2u cos(phi)/g. Always identify whether the given angle is measured from the ground or from the vertical before plugging in.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
T = 2u sin(theta)/g, where u is the launch speed, theta is the angle of projection with the horizontal, and g is the acceleration due to gravity (about 9.8 m/s^2, often taken as 10 m/s^2 in NEET).
When sin(theta) is largest, that is at theta = 90 degrees (straight up). Then T = 2u/g. For a fixed speed, a steeper launch keeps the body in the air longer, even though a 45-degree launch gives the longest horizontal range.
Both depend on the vertical component. Time to the top is t = u sin(theta)/g and maximum height is H = u^2 sin^2(theta)/(2g). Using T = 2t, you can write H = g T^2 / 8.
No. In ideal projectile motion (no air resistance), mass cancels out. A heavy ball and a light ball launched with the same speed and angle stay in the air for the same time.
Because gravity, the only force acting, is purely vertical. The horizontal velocity stays constant and has no effect on how long the body is above the ground, so it does not appear in the time-of-flight formula.
