Question

When the angle of projection is 75°, a ball falls 10 m
short of the target. When the angle of projection is 45°,
it falls 10 m ahead of the target. Both are projected
from the same point with the same speed in the same
direction, the distance of the target from the point of
projection is​

Answer 1

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Initially when ball

 

now the range of projectile is given by

R=v2sin (2θ)g

Let R be the distance of target,

R−10=v^2sin (2×75)g=v2 0.5g.......1

R+10 =v^2sin (2×45)g=v2 (1)g.........2

.2from 1

2R=3v^2/2g........3

v2g=4R/3

substitue in eq 1.

R−10=4/3R (0.5).

1/3R=10.

R=30m.

Answer 2

Answer:

The distance between the point of projection and the target will be equal to 30m.    

Explanation:

Given that a ball is projected twice towards the target from same point in the same direction with same speed.

Consider that the distance between the target and the point of projection is d.

• When the angle of projection $=75^{o}$

        The ball falls at distance $=d-10$

       ∴            $d-10=\frac{(u^{2})(sin(2*75^{o})}{g}$

       ∵      $sin150^{o}=sin(180^{o}-30^{o})=sin30^{o}=\frac{1}{2}$

        ∴            $d-10=\frac{u^{2}}{2g}$                                    ..............(1)

• When the angle of projection $=45^{o}$

       This time ball falls at distance  $=d+10$

         ∴      $d+10=\frac{(u^{2})(sin(2*45^{o})}{g}$                      ∵   $[ sin90^{o}=1]$

                  $d+10=\frac{u^{2}}{g}$                                  ................(2)

       Now divide the eq.(1) by eq.(2)

       We get     $\frac{d-10}{d+10}=\frac{1}{2}$

                     $2(d-10)=d+10\\ 2d-20=d+10$

          ∴                 $d=30m$

Therefore  there is 30m distance from the target to the point of projection.