Question

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

Answer 1

Answer:

Maximum height: 171.490 m

Horizontal Range: 391.978 m

Explanation:

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Answer 2

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, $\theta=45^{\circ}$

(i) The maximum height reached by the ball is given by :

$H=\dfrac{u^2\sin^2\theta}{2g}\\\\H=\dfrac{(62)^2\times \sin^2(45)}{2\times 9.8}\\\\H=98.06\ m$

(ii) The horizontal distance covered by the ball is called its horizontal range. It is given by :

$R=\dfrac{u^2\sin2\theta}{g}\\\\R=\dfrac{(62)^2\times \sin2(45)}{9.8}\\\\R=392.24\ m$

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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