1. A particle is projected with a speed of 25 ms-'at 30° above the horizontal. Find: (a) the time
taken to reach the highest point of the trajectory ,

1.25 s

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 25 m/s

Angle of projection (θ) = 30°

Acceleration due to gravity (g) = 10 m/s²

Time (t) to reach highest point =?

The time taken for the particle to reach the highest point can be obtained as follow:

t = uSineθ / g

t = 25 × Sine 30 / 10

t = 25 × 0.5 / 10

t = 12.5 / 10

t = 1.25 s

Thus, it will take the particle 1.25 s to reach the highest point.

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