Question

A ball is thrown such that the time of flight is 5 seconds and the horizontal range is 200m. Find the magnitude and direction of velocity with which the ball hits the ground?​

Answer 1

Answer:

$\boxed{\sf Magnitude \ of \ velocity = 5\sqrt{89} \ m/s} \\ \boxed{\sf Direction \ of \ velocity (\theta) = tan^{-1}(\frac{5}{8}) }$

Given:

Time of flight = 5 seconds

Horizontal Range = 200 m

To Find:

Magnitude and direction of velocity with which the ball hits the ground.

Explanation:

Magnitude and direction of velocity with which the ball hits the ground will be equal to magnitude and direction of velocity of projection.

Let velocity of projection be 'v'

$\sf \implies Time \ of \ flight \ (T) = \frac{2v sin \theta }{g} \\ \\ \sf \implies 5=\frac{2v sin \theta}{10} \\ \\ \sf \implies 5 = \frac{\cancel{2}v sin \theta}{\cancel{2} \times 5} \\ \\ \sf \implies 5 = \frac{v sin \theta}{ 5} \\ \\ \sf \implies \frac{v sin \theta}{ 5} = 5 \\ \\ \sf \implies v sin \theta=5 \times 5 \\ \\ \sf \implies v sin \theta= 25$

$\sf vsin \theta = v_{y} = 25 \ m/s$

$\\ \\ \sf \implies Range \ (R) =\frac{v^{2}sin2 \theta }{g} \\ \\ \sf \implies R =\frac{2v^{2}sin \theta cos \theta }{g} \\ \\ \sf \implies R =vcos \theta \times \frac{2vsin \theta }{g} \\ \\ \sf \implies R =vcos \theta \times T \\ \\ \sf \implies 200=vcos \theta \times 5 \\ \\ \sf \implies vcos \theta \times 5 = 200 \\ \\ \sf \implies vcos \theta =\frac{200}{5} \\ \\ \sf \implies vcos \theta =40$

$\sf vcos \theta = v_{x} = 40 \ m/s$

$\vec{v} = 40 \hat{i} + 25 \hat{j}$

We can calculate the magnitude of velocity as follows:

$\sf \implies v = \sqrt{v_{y}^{2} + v_{x}^{2} } \\ \\ \sf \implies v = \sqrt{(25)^{2} +(40)^{2} } \\ \\ \sf \implies v = \sqrt{625+1600} \\ \\ \sf \implies v = \sqrt{2225} \\ \\ \sf \implies v = \sqrt{25 \times 89 } \\ \\ \sf \implies v =\sqrt{5^{2} \times 89 } \\ \\ \sf \implies v = 5\sqrt{89} \ m/s$

Now, direction of the velocity can be calculate as follows:

$\sf \implies tan \theta = \frac{v_{y}}{v_{x}} \\ \\ \sf \implies tan \theta =\frac{25}{40} \\ \\ \sf \implies tan \theta = \frac{5}{8} \times\frac{\cancel{5}}{\cancel{5}} \\ \\ \sf \implies tan \theta = \frac{5}{8} \\ \\ \sf \implies \theta = tan^{-1}(\frac{5}{8} )$