Physics · Motion In A Plane · NEET
Height is a purely vertical change in position. Gravity acts only downward, so it only changes the vertical velocity, not the horizontal one. The horizontal component u cos(theta) stays constant and moves the body sideways, but it does nothing to raise or lower it. So only the vertical launch component u sin(theta) decides how high the projectile climbs, giving H = u^2 sin^2(theta) / 2g.
At the highest point the vertical velocity is zero (the body stops rising and is about to fall). But the horizontal velocity is still u cos(theta) and never becomes zero. So the total speed at the top is u cos(theta), not zero. Students who write speed = 0 at the top lose easy NEET marks. Set only the vertical part to zero when finding maximum height.
Take the vertical direction, upward positive. Initial vertical velocity = u sin(theta), final vertical velocity at top = 0, acceleration = -g, displacement = H. Substitute: 0 = (u sin(theta))^2 - 2gH. Rearrange to H = u^2 sin^2(theta) / 2g. This one-line method is faster than using the time to reach the top.
No. The formula H = u^2 sin^2(theta) / 2g has no mass in it. In the absence of air resistance, all projectiles thrown with the same speed and angle reach the same height, whether it is a bullet or a ball. Mass cancels out because gravity gives every object the same downward acceleration g.
H is largest when sin^2(theta) = 1, that is theta = 90 degrees (straight up). Then H = u^2 / 2g. Note this is different from maximum range, which happens at 45 degrees. NEET often mixes these two: maximum height needs 90 degrees, maximum range needs 45 degrees.
A bullet is fired from a gun at a speed of 280 m/s in a direction 30 degrees above the horizontal. The maximum height attained by the bullet is (g = 9.8 m/s^2, sin 30 = 0.5):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
H = u^2 sin^2(theta) / 2g, where u is the launch speed, theta is the angle above the horizontal and g is the acceleration due to gravity (about 9.8 m/s^2).
At theta = 90 degrees (thrown straight up), where H = u^2 / 2g. Do not confuse this with maximum range, which occurs at 45 degrees.
No. Only the vertical velocity is zero. The horizontal velocity u cos(theta) remains, so the speed at the top equals u cos(theta).
Time to reach the top is t = u sin(theta) / g, which is half the total time of flight. Maximum height H = u^2 sin^2(theta) / 2g can also be found from H = (1/2) g t_up^2.
Yes, in reality air resistance reduces the maximum height. But NEET problems assume no air resistance, so the ideal formula H = u^2 sin^2(theta) / 2g applies.
