A projectile is launched at an angle q above the horizontal.Three seconds later the projectile is moving the same angle q below the horizontal.Which of the following (actual values with units,not just algebraic symbols)can be found from the information given?
A)the initial vertical component of the projectile's velocity
B)the initial horizontal component of the projectile's velocity
C)the initial magnitude of the velocity
D)None of the above since at least one of the above must be given to find the other two values.

Question

Quizplus · Accepted Answer

Since the projectile is launched at an angle q above the horizontal and after 3 seconds it is moving at the same angle q below the horizontal, we can conclude that the projectile has completed half of its trajectory and is now at its maximum height. At this point, its vertical velocity is zero. Therefore, we can use the kinematic equation for the vertical motion:

Δy = v₀y × t + ½at²

where Δy is the displacement in the vertical direction, v₀y is the initial vertical component of the velocity, t is the time elapsed, and a is the acceleration due to gravity (-9.8 m/s²).

At the maximum height, Δy = 0 and t = 1.5 s (half of the total time). Therefore, we can solve for v₀y:

0 = v₀y × 1.5 + ½(-9.8)(1.5)²
v₀y = 7.35 m/s

We cannot determine the initial horizontal velocity or the initial magnitude of the velocity without additional information.

A Projectile Is Launched at an Angle Q Above the Horizontal.Three