Question

An object is projected from ground with speed
20 m/s. If horizontal range is 2 times of
maximum height, then its (Range) value is
(g = 10 m/s2)
(1)20root 5m.
(2) 8root3 m
(3) 32 m
(4) 48 m​

Answer 1

Answer:

Range value of projectile is 32mso correct option is (3)

Explanation:

we have formulas for range and maximum height of projectile as follows Range of projectile (R)

       $R= \frac{u^{2}sin 2\theta }{g}$

Maximum height of projectile : (H)  

    $H =\frac{u^{2}sin^{2} \theta }{2g}$

u = initial velocity

θ = angle of projectile

g= acceleration due to gravity

Given :  u = 20 m/s               R= 2HNow , equating           R=2H         $\frac{u^{2}sin 2\theta }{g}=2\frac{u^{2}sin^{2} \theta }{2g} \\\\ \\ sin 2\theta = sin^{2} \theta\\\\2sin \theta cos \theta = sin^{2} \theta\\\\\2cos \theta= sin \theta\\\\tan \theta = 2\\\\\theta = 63.43$

now,  to calculate range

 $R= \frac{u^{2}sin 2\theta }{g}$

 $R= \frac{20^{2}sin (2*63.43) }{10}\\\\R= 40 * 0.8\\\\R= 32 m$

Hence range of projectile is 32 m.

Answer 2

Answer:

48m

Explanation:

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