Physics
27.02.2023 05:18
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A projectile is fired directly upwards at 49.4 m/s. A second projectile is dropped

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
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Mynameismath
Mynameismath
4,6(6 marks)

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 .

IntellTanito
IntellTanito
4,8(42 marks)

A, C, E

Explanation:

Gamma rays, Microwaves, and Radio waves


Which of these are part of the electromagnetic spectrum?

Question options:
gamma rays
visible ligh

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