A projectile is fired directly upwards at 49.4 m/s.A second projectile is dropped from rest at some higher elevation at the instant the first projectile is fired and passes the first projectile 3.00 s later.From the frame of reference of the first projectile,what is the velocity of the second projectile as it passes by?
A)64.1 m/s,downward
B)71.5 m/s,downward
C)49.4 m/s,downward
D)34.7 m/s,downward
E)104 m/s,downward

Question

Quizplus · Accepted Answer

We know that the first projectile has an initial velocity of 49.4 m/s upwards. Using the kinematic equation, we can find the height it reaches in 3 seconds:

h = vit + 1/2at^2 
h = 49.4(3) - 1/2(9.8)(3)^2 
h = 44.1 m

Now, we can find the velocity of the second projectile when it passes the first projectile by considering its motion from the point it is dropped:

h = 1/2gt^2 
44.1 = 1/2(9.8)(t)^2 
t = 3 sec

The second projectile also falls for 3 seconds before passing the first projectile. Its velocity at this time can be found using the kinematic equation:

vf = vi + at 
vf = 0 + (9.8)(3) 
vf = 29.4 m/s

Since the second projectile is dropped and falls downwards, its velocity when it passes the first projectile is in the downward direction. Therefore, the answer is choice C, 49.4 m/s, downward.

A Projectile Is Fired Directly Upwards at 49