Projectile Motion (1 Viewer)

nrumble42 · May 9, 2017

1. At what angle should a particle be projected with a velocity of 50m/s to pass through a point 22.5m horizontally from and 11.25m vertically above the point of projection. Find the angle of projection and the speed of projection.

2. a stone is projected from a cliff with a speed of 30 m/s to hit an object 120m horizontally from the edge and 35m vertically below it. Find the angle of projection.

any help would be great

Mahan1 · May 9, 2017

nrumble42 said:
1. At what angle should a particle be projected with a velocity of 50m/s to pass through a point 22.5m horizontally from and 11.25m vertically above the point of projection. Find the angle of projection and the speed of projection.

2. a stone is projected from a cliff with a speed of 30 m/s to hit an object 120m horizontally from the edge and 35m vertically below it. Find the angle of projection.

any help would be great

$\textrm{I am going to do the first part in details then the second part is going to be exactly the same.}$

$\textrm{ Write out the equations of the motion:}$

$x = 50 \ cos\theta \ t , y = 50\ sin\theta \ t - \frac{1}{2} g \ t^{2} = 50\ sin\theta \ t - 4.9 \ t^{2} \ \textrm{where g =9.8}$

$\textrm{in our case,} 22.5 = 50 \ cos\theta \ t, 11.25 = 50 \ sin\theta \ t - 4.9 \ t^{2} \ \ \ (*)$

$\textrm{From the first equation we get } t = \frac{22.5}{50 \ cos \theta } \ \textrm{sub it into the second equation:}$

$11.25 = 50 \ sin\theta \ \frac{22.5}{50 \ cos \theta } - 4.9 \times (\frac{22.5}{50 \ cos \theta } )^{2}$

$\Rightarrow 11.25 = \frac{22.5 \ sin\theta}{cos\theta} - 4.9 \times (\frac{22.5}{50})^{2} sec^{2}\theta$

$\textrm{Using the following trig identity, we can simplify the above equation: }$
$tan^{2}\theta + 1 = sec^{2}\theta$

$\Rightarrow 11.25 = 22.5 \ tan \theta - 4.9 \times (\frac{22.5}{50})^{2} sec^{2}\theta = 22.5\ tan\theta - 4.9 \times (\frac{22.5}{50})^{2} (1 + tan^{2}\theta )$

$\Rightarrow 11.25 + 4.9 \times (\frac{22.5}{50})^{2} = 22.5\ tan\theta - 4.9 \times (\frac{22.5}{50})^{2} tan^{2}\theta$

$12.24225 = 22.5 \ x - 0.99225 \ x^{2} \ \textrm{where} \ x = tan\theta$

$\textrm{Solving the quadratic equation, yields} \ x = \frac{22.5 \pm \sqrt{22.5^{2} - 4 \times 0.99225 \times 12.24225}}{2 \times 0.99225}$

$\Rightarrow x = 22.11791, 0.55782 \ \textrm{Since} \ x = tan\theta \Rightarrow \theta = 87.41, 29.15$

$\textrm{subbing the values of } \theta \ \textrm{into} \ (*) \ \textrm{we obtain }\theta = 29.15 \ \textrm{is the only answer}$

$\textrm{For the second one, follow the same reasoning but watch out the y in this case is } -35.$

$\textrm{Also you get two different answers and both of them are correct}$

Projectile Motion (1 Viewer)

nrumble42

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Mahan1

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