Height and DistanceHard
Question
From the top of a 64 metres high tower, a stone is thrown upwards vertically with the velocity of 48 m/s. The greatest height (in metres) attained by the stone, assuming the value of the gravitational acceleration g = 32 m/s2, is :
Options
A.128
B.88
C.112
D.100
Solution
At maximum height v = 0
Now v2 = u2 − 2gh
⇒ 0 = (48)2 − 2(32)h.
⇒ h = 36
Maximum height = 36 + 64 = 100 mt
Now v2 = u2 − 2gh
⇒ 0 = (48)2 − 2(32)h.
⇒ h = 36
Maximum height = 36 + 64 = 100 mt
Create a free account to view solution
View Solution FreeMore Height and Distance Questions
PQR is a triangular park with PQ = PR = 200 m. A T.V. tower stands at the mid-point of QR. If the angles of elevation of...At a point 15 m away from the base of a 15 m high house the angle of elevation of the top is -...Let 10 vertical poles standing at equal distances on a straight line, subtend the same angle of elevation at apoint O on...If the angles of elevation of the top of a tower from three collinear points A, B and C, on a line leading to the foot o...From the top of a cliff 25 meters high, the angle of elevation of a tower is found to be equal to the angle of depressio...