A particle is projected vertically upwards with a speed of 40 m/s. At what times will it be 35m above its point of projection? (text{Take g} = 10 ms^{-2})
-
A.
1 sec and 7 sec -
B.
1 sec and 8 sec -
C.
2 sec and 5 sec -
D.
2 sec and 7 sec
Correct Answer: Option A
Explanation
(s = ut + frac{1}{2}at^{2})
(s = ut – frac{1}{2}gt^{2}) (Upward movement against gravity)
(35 = 40t – frac{1}{2}10t^{2})
(35 = 40t – 5t^{2})
(5t^{2} – 40t + 35 = 0)
(t^{2} – 8t + 7 = 0)
((t – 1)(t – 7) = 0 implies t = text{1 sec and 7 sec})