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Please explain the image problem fully not only answer I need to be explained......... Please........,.,..,.,.
Height of the cliff, h = 240 m Velocity of Stone 1, y₁= 10 m/s Velocity of Stone 2, y₂ = 40 m/s To calculate the time when stone 1 strikes the ground, use Equation of Motion, h = ut + ½ gt² -240 = 10 × t – ½ × 10 × t² 5t² -10t -240 = 0 t² – 2t – 48 = 0 t = 8 sec and -6 sec To calculate the time when stone 2 strikes the ground, use Equation of Motion, h = ut + ½ gt² -240 = 40 × t – ½ × 10 × t² 5t² -40t -240 = 0 t = 12 sec The relative velocity of the two stone = y₂ – y₁ = 40 – 10 = 30 m/s Therefore, graph of y₂ – y₁ against time is a straight line. Hence, when the stone strikes the ground, the relative velocity will increase up to 12 seconds as shown in graph (3).
Height of the cliff, h = 240 m
Velocity of Stone 1, y₁= 10 m/s
Velocity of Stone 2, y₂ = 40 m/s
To calculate the time when stone 1 strikes the ground, use Equation of Motion,
h = ut + ½ gt²
-240 = 10 × t – ½ × 10 × t²
5t² -10t -240 = 0
t² – 2t – 48 = 0
t = 8 sec and -6 sec
To calculate the time when stone 2 strikes the ground, use Equation of Motion,
-240 = 40 × t – ½ × 10 × t²
5t² -40t -240 = 0
t = 12 sec
The relative velocity of the two stone = y₂ – y₁ = 40 – 10 = 30 m/s
Therefore, graph of y₂ – y₁ against time is a straight line.
Hence, when the stone strikes the ground, the relative velocity will increase up to 12 seconds as shown in graph (3).
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