A student drops a ball from the top of a tall building; it takes 2.9 s for the ball to reach the ground.
1 answer:
<h2><u>We have</u>,</h2>
- Initial velocity (u) = 0 m/s
- Time taken (t) = 2.9s
- Acceleration due to gravity (g) = + 10 m/s² [Down]
- Final velocity (v)
- Height (h)
→ v = u + gt
→ v = 0 + 10(2.9)
→ v = 29 m/s … ( Ans )
And,
→ h = ut + ½gt²
→ h = 0(2.9) + ½ × 10 × (2.9)²
→ h = 5 × 8.41
→ h = 42.05 m … ( Ans )
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Answer:
The mass of the banana is m and it is at height h.
Applying the Law of Conservation of Energy
Total Energy before fall = Total Energy after fall
=
Here, total energy is the sum of kinetic energy and potential energy
+
=
+
(a)
When banana is at height h, it has
= 0 and
= mgh
and when it reaches the river, it has
= 1/2m
and
= 0
Putting the values in equation (a)
0 + mgh = 1/2m + 0
mgh = 1/2m
<em>cutting 'm' from both sides</em>
<em> </em>gh = 1/2
v =
Hence, the velocity of banana before hitting the water is
v =