An object is thrown upwards with an initial velocity of 17m/sec from a building with 12m height. It is at a height of S=12+17t-5t^2 from the ground after a flight of 't' seconds. Find the time taken by the object to touch the ground?
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S=12+17t-5t^2 [eqn 1]
Initial velocity u= 17m/s
Velocity at any time t, V= dS/dt [I.e differentiate S with respect to t, since V=d/t]
Therefore V =17 - 10t
At max height V=0
Therefore time to reach max height
T=17/10=1.7s
To get distance covered in 1.7s, put T=1.7s in equation 1
S=26.45m
Total height from ground level = 12+26.45=38.45m
Using eqn 1 again, put S=38.45 and calculate for t
38.45= 12+17t-5t^2
Solve the simultaneous equation
5t^2 - 17t + 26.45
to get t
Initial velocity u= 17m/s
Velocity at any time t, V= dS/dt [I.e differentiate S with respect to t, since V=d/t]
Therefore V =17 - 10t
At max height V=0
Therefore time to reach max height
T=17/10=1.7s
To get distance covered in 1.7s, put T=1.7s in equation 1
S=26.45m
Total height from ground level = 12+26.45=38.45m
Using eqn 1 again, put S=38.45 and calculate for t
38.45= 12+17t-5t^2
Solve the simultaneous equation
5t^2 - 17t + 26.45
to get t
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7
Answer: Time taken by the object to touch ground is 4 seconds
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