A ball is thrown from ground level of a field with a speed of 12-0 m - s at an angle of 45- with the horizontal- At what distance will it hit the field again- Take g-10-0 m - s 2

The horizontal range = u^2 sin 2θ/g = (12 m/s)^2 × sin (2 × 45^∘)/10m/s^2 = 144 m^2/s^2/10.0m/s^2 = 14.4m Thus, the ball hits the field at 14.4 m from the point ...

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Standard XII

Physics

Range

A ball is thr...

Question

A ball is thrown from ground level of a field with a speed of 12.0 m/s at an angle of $45^{\circ}$ with the horizontal. At what distance will it hit the field again? Take g = 10.0 m/ $s^{2}$ .

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Solution

The horizontal range = $\frac{u^{2} s i n 2 θ}{g}$

= $\frac{(12 m / s)^{2} \times s i n (2 \times 45^{\circ})}{10 m / s^{2}}$

= $\frac{144 m^{2} / s^{2}}{10.0 m / s^{2}}$ = 14.4m

Thus, the ball hits the field at 14.4 m from the point of projection.

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A ball is thrown from ground level of a field with a speed of 12.0 m/s at an angle of 45∘ with the horizontal. At what distance will it hit the field again? Take g = 10.0 m/s2. ___

The horizontal range = u2sin2θg = (12m/s)2×sin(2×45∘)10m/s2 = 144m2/s210.0m/s2 = 14.4m Thus, the ball hits the field at 14.4 m from the point of projection.

A ball is thrown from ground level of a field with a speed of 12.0 m/s at an angle of $45^{\circ}$ with the horizontal. At what distance will it hit the field again? Take g = 10.0 m/ $s^{2}$ .

___

The horizontal range = $\frac{u^{2} s i n 2 θ}{g}$

= $\frac{(12 m / s)^{2} \times s i n (2 \times 45^{\circ})}{10 m / s^{2}}$

= $\frac{144 m^{2} / s^{2}}{10.0 m / s^{2}}$ = 14.4m

Thus, the ball hits the field at 14.4 m from the point of projection.