A ball is thrown into air with a speed of 62 m/s at an angle of 45° with the horizontal. Calculate
i) maxim height
ii) Horizontal range
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A ball is thrown into air with a speed of 62 m/s at an angle of 45° with the horizontal. Calculate
i) maxim height
ii) Horizontal range
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Answer:
Maximum height: 171.490 m
Horizontal Range: 391.978 m
Explanation:
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The maximum height of the ball is 98.06 m and the horizontal range is 392.24 m.
Explanation:
Given that,
Initial speed of the ball, u = 62 m/s
Angle of projection with the horizontal, [tex]\theta=45^{\circ}[/tex]
(i) The maximum height reached by the ball is given by :
[tex]H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(62)^2\times \sin^2(45)}{2\times 9.8}\\\\H=98.06\ m[/tex]
(ii) The horizontal distance covered by the ball is called its horizontal range. It is given by :
[tex]R=\dfrac{u^2\sin2\theta}{g}\\\\R=\dfrac{(62)^2\times \sin2(45)}{9.8}\\\\R=392.24\ m[/tex]
So, the maximum height of the ball is 98.06 m and the horizontal range is 392.24 m.
Learn more,
Projectile motion
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