Question

3. An object of mass 1kg was dropped from a tower. When the object was at half the height of the tower, its kinetic energy was 700J. What is the height of the tower (take g = 10 m/s^2). How long has the object been falling?​

Answer 1

FINDING VELOCITY AT HALF THE HEIGHT.

Kinetic energy of an object =¹/₂mv²

We have been given: Mass (m) = 1kg

K.E = 700J

Using the above equation,

¹/₂mv² = 700

=> v² = ⁷⁰⁰ ˣ ²/ ₘ

=> v²= 1400 metres²

=> v = √1400

= 37.42 ms⁻¹

ANALYSIS:-

When the ball was thrown, it's initial velocity was 0 and when it covered the distance ʰ/₂ ( h =height of tower) its velocity was 37.42 ms⁻¹

And when the ball fell, it did so under the effect of gravity. So its acceleration was g= 9.8 ms⁻¹

CALCULATING HEIGHT OF THE TOWER:

Using v² - u² = 2ax , where x = distance covered, We get,

x = (ᵛ² ⁻ ᵘ²) /₂ₐ

= (37.42 -0) / 2(9.8)

= 1.91 metres

CALCULATING TIME TAKEN FOR FALL:

Using v =u +at we get,

37.42 = 9.8(t)

=> t = 3.82 seconds