A stone is thrown vertically upwards with a velocity of 40m/s and is caught back. Taking g= 10m/s, calculate the maximum height reached by the stone. What is the net displacement and total distance covered by the stone?
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2
displacement =0
when the velocity became 0 at the point it will be height
when the velocity became 0 at the point it will be height
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A = 10 M/S²
U = 40 M/S
V = 0 M/S
USING THE 3RD EQUATION OF UNIFORMLY ACCELERATED MOTION,
V² = U² + 2GH
0 = 40² + 2×-10×H
-400 = -20H
H = 20 METRES.
MAXIMUM HEIGHT REACHED BY THE STONE WILL BE 20M.
NET DISPLACEMENT = 0 BECAUSE THE STONE COMES BACK FROM WHERE IT WAS THROWN.
TOTAL DISTANCE WILL BE 2 X H
THAT IS,
TOTAL DISTANCE = 40 METRES
PLZ MARK IT AS BRAINLIEST ANSWER AND DROP A ♥
U = 40 M/S
V = 0 M/S
USING THE 3RD EQUATION OF UNIFORMLY ACCELERATED MOTION,
V² = U² + 2GH
0 = 40² + 2×-10×H
-400 = -20H
H = 20 METRES.
MAXIMUM HEIGHT REACHED BY THE STONE WILL BE 20M.
NET DISPLACEMENT = 0 BECAUSE THE STONE COMES BACK FROM WHERE IT WAS THROWN.
TOTAL DISTANCE WILL BE 2 X H
THAT IS,
TOTAL DISTANCE = 40 METRES
PLZ MARK IT AS BRAINLIEST ANSWER AND DROP A ♥
Akv2:
:)
☆☆☆
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