Physics

Problem

A cricket ball is struck from ground level with a speed of

25\,\text{m\,s}^{-1}

at an angle of

37^{\circ}

to the horizontal. Take

g = 9.8\,\text{m\,s}^{-2}

. a. (i) Calculate the horizontal and vertical components of the initial velocity. b. (ii) Determine the horizontal position

x

and the vertical position

y

of the ball after

2\,\text{s}

from launch. c. (iii) Find the magnitude and direction (angle with the horizontal, measured above or below the horizontal as appropriate) of the velocity of the ball at

t = 2\,\text{s}

. Round your final answers to two significant figures.

Intuitive explanation

Projectile motion splits into two independent stories: horizontal motion at constant velocity, and vertical motion under gravity. The key is to resolve your initial velocity into components using the launch angle, then use those to find time of flight, range, and maximum height. The angle tells you how much of that 25 m/s goes "forward" versus "up"—and only the vertical part fights gravity.

Intuitive explanation

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