Question

25. A projectile is fired from the top of a tower of height 20 m with an
initial speed of 25 m/s at an unknown angle. On reaching the
ground its speed would be​

Answer 1

Answer:

Explanation:

A projectile is fired from the top of a 40 m high cliff with an initial speed of 50 m/s at an unknown angle. Find its speed when it hits the ground.

Answer 2

Answer:

32 m/s

Explanation:

Maximum height reached by the projectile = u²/2g

= 25²/2 × 10

= 625/20

= 31.25 m

The ball returns back to the ground.

Total height = 31.25 + 20

= 51.25 m

Initial velocity = 0 as it is now a freely falling body.

Now we should use the 3rd equation of motion

= v² - u² = 2gh

Since u = 0.

v² = 2gh

v² = 2 × 10 × 51.25

v² = 1025

v = √1025 = 32.01 ≈ 32 m/s