Q#32 (Past Exam Paper – March 2019 Paper 12 Q7)
A stone is thrown vertically upwards from a point that is 12 m above the sea. It then falls into the sea below after 3.4 s.
Air resistance is negligible.
At which speed was the stone released when it was thrown?
A 3.5 m s-1 B 6.6 m s-1 C 13 m s-1 D 20 m s-1
Solution:
Answer: C.
The motion of the stone consists of both an upward and a downward movement. So, we need to be careful with the signs of the vector quantities involved.
Let the upward direction be positive. Any quantity pointing downwards would be negative.
Initial velocity = u (???)
Take the point at which the stone is released to be the origin.
The sea is at a distance of 12 m below the origin. So, the displacement would be negative.
Displacement s = – 12 m
Acceleration a = – 9.8 m s-2 (this is also negative as it is downwards)
Time t = 3.4 s
Consider the equation of uniformly accelerated motion:
s = ut + ½ at2
- 12 = 3.4u + (½ × -9.8 × 3.42)
- 12 = 3.4u = 3.4u – 56.644
3.4u = -12 + 56.644
u = 44.644 / 3.4
Initial speed u = 13.1 m s-1
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