Physics · Motion In A Straight Line · NEET
Velocity is zero at the highest point, but acceleration is NOT zero. Gravity still pulls the ball down, so a = -g = -10 m/s^2 (or -9.8) at every instant, including the top. On the v-t graph this is the point where the line crosses the time axis (v = 0), but the line keeps its same downward slope through it. This is the single most tested trap in this topic.
Because the acceleration is constant (only gravity acts, a = -g). A constant acceleration means velocity changes by the same amount every second, so v = u - gt is a straight line. Its slope is -g. A curved v-t graph would mean changing acceleration, which does not happen in free fall.
It is a downward-opening parabola, like an upside-down U or a hill. Height increases at a decreasing rate, reaches a maximum (the top of the flight) where the slope is zero, then decreases. The slope of the x-t graph at any instant equals the velocity, so the slope is positive going up, zero at the top, and negative coming down.
Below the axis means velocity is negative, which just means the ball is now moving downward (we chose up as positive). On the way up v is positive; after the top the ball moves down so v becomes negative. When it returns to the thrower's hand, v = -u, equal in size to the launch speed but opposite in direction.
If up is positive, acceleration is -g the WHOLE flight, both going up and coming down. On the way up it slows the ball (opposite to motion); on the way down it speeds the ball (same as motion). The sign of g never changes just because the ball reverses direction; only the sign of velocity changes.
A ball is thrown vertically upward and falls back to the thrower. Which velocity (v)-time (t) graph correctly represents its motion? (upward taken positive)
A ball is thrown vertically upward with a velocity of 4 m/s from a bridge. The ball strikes the water surface after 4 s. The height of the bridge above the water surface is (g = 10 m/s^2):
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The slope equals the acceleration, which is -g (about -9.8 m/s^2, or -10 m/s^2 if g = 10 is given). It is constant, negative, and the same on the way up and the way down.
No. The line only touches v = 0 at one instant (the top) and keeps its same negative slope. It never becomes flat, because acceleration is never zero during the flight.
Time to go up equals time to come down (t = u/g each), and the motion has constant acceleration, so the height-time parabola is symmetric about the vertical line through the top point.
Area under the v-t graph gives displacement. The positive area (going up) and the negative area (coming down) are equal when the ball returns to the start, so the net displacement is zero even though distance travelled is 2H.
Check three things: acceleration is constant so the v-t graph must be a straight line (not curved); up-positive means v must cross into negative values; and the slope must be negative (-g). Only the straight line crossing v = 0 fits all three.
