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Home / Doubts and Solutions/JEE/Physics

A stone is thrown from the top of a building upward at an
angle of 30.0° to the horizontal and with an initial speed of
20.0 m/s. If the height of the building
is 45.0 m, (a) how long is it before the stone hits the
ground?

Asked by gargpuneet989 26th June 2018, 12:11 PM
Answered by Expert
Answer:
For conveneience of calculation, we use g = acceleration due to gravity = 10 m/s2
vertical component of projection velocity = 20×sin30 = 10 m/s
 
let t1 be the time taken by the stone to reach maximum height
time taken to reach maximum height is calculated using the formula " v = u - g×t ", with final speed v =0, initial speed u = 10 m/s.
 
t1 = u/g = 10/10 = 1 s.
 
Let h be the maximum height reached by the stone
maximum height is calculated using the formula " S = u×t-(1/2)×g×t2 ", with initial velocity u = 10 m/s , S = h and time of travel t =1.
 
h = 10×1 -(1/2)×10×1×1 = 5 m.
 
Since building height is 45 m, total vertical distance to be travelled to reach ground is 45+5 = 50 m.
 
time t2 to reach ground from maximum height is calculated using formula " S = u×t+(1/2)×g×t2 " , with initial velocity u = 0 and S = 50 m.
 
50 = (1/2)×10×t22
solving the above eqn for t2, we get t2 =√10 .
Hence total time = t1 + t2 = √10+1 = 4.16 s
Answered by Expert 27th June 2018, 10:47 AM
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