Question

12)
A ball is thrown obliquely so that it just clears horizontally a tree 25m high and at a horizontal distance of 50m
from the point of projection. Find the direction and speed of projection of the ball.
[Ans/Elevation of 45° to the horizontal; 31.3 m/s]​

Answer 1

Answer:

Since,it clears wall high when it is moving horizontally 3 meter is its maximum height.

⇒2gu2sin2θ=3m

Also, u=20m/s

⇒20202sin2θ=3

⇒sin2θ=203

⇒θ=sin−1203

=22.780

Answer 2

The angle of projection from horizontal is equal to 45° and the speed of projection is equal to 31.62 ms⁻¹.

Given:

The horizontal distance covered by the ball to reach the tree = 50 m

The ball just clears a tree of height equal to 25 m.

To Find:

The direction and speed of projection.

Solution:

Let the angle of projection from the horizontal be 'θ'

Let the speed of projection be 'v'

Assume: Acceleration due to gravity (g) = 10 ms⁻²

→ The ball just clears a tree of height equal to 25 m, this means that the maximum height of projection (H) is equal to 25 m.

               ∴ Maximum height of projection (H) = 25 m

→ The formula for the maximum height of projection (H) is given as:

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

     → where v = speed of projection

                    θ = angle of projection

                                    $H=\frac{v^{2} sin^{2}\theta }{2g}=25\\\\\therefore v^{2} sin^{2}\theta = 50g = 500-(i)$

→ The tree is situated at the position of the maximum height of the projection which is half of the range of projection (R).

∵ The ball covers a horizontal distance of 50 m while reaching the tree.

∴ The range of the projection (R) would be double of 50 m.

∴ Range of projection (R) = 2 × 50 = 100 m

→ The formula for the range of projection (R) is given as:

                                      $R=\frac{v^{2} sin2\theta}{g}$

     → where v = speed of projection

                    θ = angle of projection

                                       $R=\frac{v^{2} sin2\theta}{g}=100\\\\\therefore v^{2} sin2\theta=100g=1000\\\\\therefore 2v^{2} sin\theta cos\theta=1000\\\\\therefore v^{2} sin\theta cos\theta=500-(ii)$

→ Dividing equation (i) by equation (ii):

                                    $\frac{Eq.(i)}{Eq.(ii)} :\frac{v^{2}sin^{2} \theta }{v^{2} sin\theta cos\theta} =\frac{500}{500}$

                                                $\therefore \frac{sin\theta}{cos\theta} =1\\\\\therefore tan\theta = 1\\\\\therefore \theta=45^0$

→ Hence the angle of projection from horizontal (θ) comes out to be equal to 45°

→ Now putting the value of 'θ' in equation (i):

                                            $v^{2} sin^{2} \theta=500$

                                         $v^{2} sin^{2} 45^0 = 500$

                                                $\frac{v^{2} }{(\sqrt{2})^2 }= 500$

                                                    $\frac{v^{2} }{2} =500\\\\ v^{2} =1000$

                                                  $\therefore v=\sqrt{1000}\\\\ \therefore v=31.62ms^{-1}$

→ Hence the speed of projection (v) comes out to be equal to 31.62 ms⁻¹.

Therefore the angle of projection from horizontal (θ) is equal to 45° and the speed of projection (v) is equal to 31.62 ms⁻¹.

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