Conceptual Problem about Projectile Motion| University Physics

The figure shows the trajectory (i.e., the path) of a ball undergoing projectile motion over level ground. The time t_0=0\:s corresponds to the moment just after the ball is launched from position x_0=0\:m and y_0=0\:m. Its launch velocity, also called the initial velocity, is \vec{v}.

Two other points along the trajectory are indicated in the figure.

  • One is the moment the ball reaches the peak of its trajectory, at time t_1 with velocity \vec{v}_1. Its position at this moment is denoted by \left(x_1,\:y_1\right) or \left(x_1,\:y_{max}\right) since it is at its maximum height.

  • The other point, at time t_2 with velocity \vec{v}_2, corresponds to the moment just before the ball strikes the ground on the way back down. At this time its position is \left(x_2,\:y_2\right), also known as \left(x_{max},\:y_2\right) since it is at its maximum horizontal range.

Projectile motion is symmetric about the peak, provided the object lands at the same vertical height from which it was launched, as is the case here. Hence y_2=y_0=0\:m.

The parabolic trajectory is shown on the xy plane. The trajectory starts at the origin with coordinates labeled as x 0 and y 0. The initial velocity of the ball is v 0, and it is directed upward and to the right. The initial moment of time is t 0. The object rises to a maximum height y 1 or y max. At this moment of time labeled as t 1, the velocity v 1 is horizontal and directed to the right, and the position of the ball is x 1. Then the ball falls to the ground. It lands at a point that is at distance x 2 or R from the origin. At this moment of time labeled as t 2, the velocity v 2 is directed downward and to the right.

How do the speeds v_0,\:v_1,\:and\:v_2 compare?

ANSWER: v_0=v_2>v_1>0

Here v0 equals v2 by symmetry and both exceed v1. This is because v0 and v2 include vertical speed as well as the constant horizontal speed.