A body is thrown vertically upwards with an initial velocity of 9.8 ms⁻¹. What is its speed and direction after (i) 1s (ii) 2s (g = 9.8 ms⁻²)?
Also, find the height to which it will rise.
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Answered by
1
(1) 0 meter/second(2) 9.8meter/second
Answered by
2
Maximum height to which the stone can reach is 4.9 m.
Explanation:
As the stone is thrown upward the acceleration due to gravity is to be taken negative.
g = - 9.8 m/s2
Given that initial velocity u = 9.8 m/s
At the highest point, final velocity will be zero. v = 0
v2 - u2 = 2as
Substituting, we get:
0 - (9.8)2 = 2× (-9.8) × h
- 9.8 / 2 = h
h = 4.9 m
So, the maximum height to which the stone can reach is 4.9 m.
Time take for the object to fall to ground will be:
s = ut + 1/2at^2
u will be 0 and s is 4.9m with g being positive.
4.9 = 0 + 1/2*9.8*t^2
t = 1 seconds
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