Example 13 Time and Projectile Motion

A projectile is launched from and returns to ground level, as Figure 3.22 shows. Air resistance is absent. The horizontal range of the projectile is R=175 m, and the horizontal component of the launch velocity is v_0x=25 m/s. Find the vertical component v_0y of the launch velocity.

A projectile, launched with a velocity whose horizontal component is

, has a range of

. From these data the vertical component v
0y
of the initial velocity can be determined.

Figure 3.22 A projectile, launched with a velocity whose horizontal component is

, has a range of

. From these data the vertical component v _0y of the initial velocity can be determined.

What is the final value of the horizontal component v_x of the projectile’s velocity?

The final value v_x of the horizontal component of the projectile’s velocity is the same as the initial value in the absence of air resistance. In other words, the horizontal motion occurs at a constant velocity of 25 m/s.

Can the time be determined for the horizontal part of the motion?

Yes. In constant-velocity motion, the time is just the horizontal distance (the range) divided by the magnitude of the horizontal component of the projectile’s velocity.

Is the time for the horizontal part of the motion the same as the time for the vertical part of the motion?

Yes. The value for the time calculated for the horizontal part of the motion can be used to analyze the vertical part of the motion.

For the vertical part of the motion, what is the displacement of the projectile?

Since the projectile is launched from and returns to ground level, the vertical displacement is zero.

Solution From the constant-velocity horizontal motion, we find that the time is

For the vertical part of the motion, we know that the displacement is zero and that the acceleration due to gravity is –9.80 m/s², assuming that upward is the positive direction. Therefore, we can use Equation 3.5b to find the initial y component of the velocity: