Question

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

Answer 1

Answer:

Explanation:

velocity of first projectile after 3 s

v = u - gt

v = 49.4 - 9.8 x 3

= 20 m /s

Velocity of second projectile after 3 s after being dropped from rest

v = u + gt

= 0 + 9.8 x 3

= 29.4 m /s

They will be moving in opposite direction at the time of meeting , so their relative velocity

= 20 + 29.4 = 49.4 m /s

From the frame of reference of the first projectile, the velocity of the second projectile will be 49.4 m /s .