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- Consider a body P
**projected vertically**upwards from the surface of the earth with an initial velocity, ... For the motion of a particle**projected vertically up**or down from some**height**(**h**) ... Referring to the previous Problem , if the**stone reaches**the ground after t = 2 second from same initial**height**of release, find the (a) ... **A stone is projected vertically**upward from the top of a 120-m high tower at a initial velocity of 36 m/s. What is the vertical distance between the**stone**and the ground after t = 9 seconds? ... vh=25 m/s**Height h**= 60 m When the**stone is projected**...Time**to reach maximum height**t = (uSinθ)/g = (3.13×Sin30°). Answer (1 of 3): Given expression for**height**is**h**= 60t-5t^2 Also the required ...- Question: A ball thrown
**vertically**upwards after reaching a**maximum height h**, returns to the starting point after a time of l0 s. Its displacement is. a) zero; b) 10**h**; c) 2**h**; d)**h**; Answer: zero Question: The ball**is projected up**from ground with speed 30 m/sec. **A stone is projected vertically**upward from the top of a 120-m high tower at a initial velocity of 36 m/s. What is the vertical distance between the**stone**and the ground after t = 9 seconds? ... vh=25 m/s**Height h**= 60 m When the**stone is projected**...Time**to reach maximum height**t = (uSinθ)/g = (3.13×Sin30°). Answer (1 of 3): Given expression for**height**is**h**= 60t-5t^2 Also the required ...**A stone projected**with a velocity u at an angle θ with the horizontal**reaches maximum height**H1. When it**is projected**with velocity u at an angle π2−θwith the horizontal, it**reaches maximum height**H2. The relation between the horizontal range R of the projectile, H1 and H2 is (1) R=4H1H2 (2) R = 4(H1 – H2) (3) R = 4(H1 + H2) (4) R=H12H22 Practice questions, MCQs, Past.