Question

A ball is thrown with a velocity of 15m/s at an angle of 30degree with the horizontal determine (i) time of the flight of the ball (ii) maximum height attained by the ball (g =10 m/s)

Answer 1

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Answer 2

Given:

Initial velocity, u = 15 m/s

Angle with the horizontal, α = 30°

To Find: Time of flight and maximum height

Calculation:

Time of flight of projectile motion is the total time taken by a ball to reach the ground. It is given as:

$T=\frac{2usin\alpha }{g} \\T=\frac{2*15*sin (30)}{10} \\T=\frac{15}{10} =1.5s$

Maximum height is the maximum distance from the ground that the ball travels in air.

$H=\frac{u^{2} sin^{2} \alpha }{2g} =\frac{15*15*0.5*0.5}{20} \\H=2.81 m$

Answer:

Maximum height of the ball is 2.81 metres.

Time of flight of the ball is 1.5 seconds.