A ball is thrown with an initial velocity of 20 m/s at an angle of 60° above the horizontal. If we can neglect air resistance, (a) what is the horizontal component of its instantaneous velocity at the exact top of its trajectory? (b) how long is the ball in flight before it hits the ground?(c) how far has the ball traveled in the horizontal direction when it lands?

[tex]s=vsin \theta t+\frac{1}{2}at^2\\\Rightarrow 0=20sin60+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2\times 20sin 60}{9.81}}\\\Rightarrow t=1.87914\ s[/tex]

The time the ball is in the air is 1.87914 s

Range is given is by

[tex]R=vcos\theta t\\\Rightarrow R=10\times 1.87914\\\Rightarrow R=18.7914\ m[/tex]

The range is 18.7914 m

onyebuchinnaji onyebuchinnaji · Answer 2 · 2021-10-01T20:05:32+02:00

(a) the horizontal component of its instantaneous velocity at the exact top of its trajectory is 10 m/s.

(b) The time of flight of the object is 3.53 s

(c) The horizontal distance traveled by the ball when it lands is 35.35 m.

The given parameters;

initial velocity of the ball, u = 20 m/s

angle of projection, Ф = 60°

(a) The horizontal component of its instantaneous velocity at the exact top of its trajectory.

the initial horizontal component of the velocity;

[tex]v_0_x = v_0 \times cos(\theta)\\\\v_0_x = 20 \times cos(60)\\\\v_0_x = 10 \ m/s[/tex]

At the top of its trajectory, the velocity is equal to its initial value.

Thus, the horizontal component of its instantaneous velocity at the exact top of its trajectory is 10 m/s.

(b) The time of flight of the object is calculated as;

[tex]T = \frac{2usin(\theta)}{g} \\\\T = \frac{2\times 20\times sin(60)}{9.8} \\\\T = 3.53 \ s[/tex]

(c) The horizontal distance traveled by the ball when it lands;

[tex]R = \frac{u^2 sin(2\theta)}{g} \\\\R = \frac{20^2 \times sin(120)}{9.8} \\\\R = 35.35 \ m[/tex]

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