Physics, 12.08.2020 08:01 jay555538

A projectile has a range of 60 m and can reach a maximum height of 15 m. Calculate the angle at which projectile is fired? ( g = 10m/s2 )

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Answer from: stodd9503

The angle the projectile was fired is 45^o

Explanation:

Recall the formulas for maximum height and ranges for a projectile fired with initial velocity "v" at an angle $\theta$ :

$h = \frac{v^2\,sin^2(\theta)}{2\,g}\\R=\frac{v^2\,sin(2\,\theta)}{g}$

we can use them to solve for the angle by first isolating the value v^2 which is common in both equations:

$v^2=2\,h\,g/sin^2(\theta)=2\,(15)\,g/sin^2(\theta)=30\,(g)/sin^2(\theta) \\ \\v^2=R\,g/sin(2\,\theta)=60\,(g)/sin(2\,\theta)$

and now, making those to expressions equal and using the formula for the sine of a double angle, we get:

$\frac{30\,(9.8)}{sin^2(\theta)} =\frac{60\,(g)}{sin(2\,\theta)} \\30\,(g)\,sin(2\,\theta)=60\,(g)\,sin^2(\theta)\\sin(2\,\theta)=2\,sin^2(\theta)\\2\,sin(\theta)\,cos(\theta)=2\,sin(\theta)\,sin(\theta)\\cos(\theta)=sin(\theta)$

This happens when $\theta=45^o$

Answer from: Quest

here we know that ball will drop down by total distance 3.2 m

so here we know that

$v_{yi} = 0$

$\delta y = - 3.2 m$

now using kinematics

$\delta y = v_{yi} t + \frac{1}{2}at^2$

$-3.2 = 0 -\frac{1}{2}(9.8) t^2$

$t = 0.81 s$

now in the same time it will cover horizontal distance of 0.75 m

$\delta x = 0.75m$

so speed of the ball will be

$v = \frac{delta x }{\delta t}$

$v = \frac{0.75}{0.81} = 0.93 m/s$

Answer from: Quest

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A projectile has a range of 60 m and can reach a maximum height of 15 m. Calculate the angle at whic...

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