Question 72015PhysicsWAEC

A body is projected with an initial velocity u at an angle θ to the horizontal. The time taken by it to reach its maximum height is given by the expression

A

2usin⁡θg\frac{2u\sin\theta}{g}

B

u2sin⁡2θg\frac{u^2\sin^2\theta}{g}

C

usin⁡θg\frac{u\sin\theta}{g}

D

usin⁡θ2g\frac{u\sin\theta}{2g}

Step-by-Step Explanation

The time taken to reach maximum height in projectile motion is determined by the vertical component of the initial velocity, which is given by usin⁡θu\sin\theta. At maximum height, the vertical velocity becomes zero, and using the equation of motion v=u+atv = u + at, where v=0v = 0 and a=−ga = -g, leads to the expression t=usin⁡θgt = \frac{u\sin\theta}{g}. This formula reflects the relationship between the initial vertical velocity, gravitational acceleration, and time.

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