Question 32
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
Reference: Past Exam Paper – March 2019 Paper 12 Q7
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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