Max horizontal displacement of a projectile

[Phys student]

Homework Statement

A catapult is capable of firing a projectile with a velocity of 25 m/s, the catapult is surrounded by a wall of height 20.4 m what is the maximum distance measured from the wall, that the catapult
projectiles are capable of hitting?

Homework Equations

h=(v_o^2sin^2θ)/2g
x=v_otcosθ
y=v_otsinθ -0.5gt^2
R=v_ocosθ/g * (v_osinθ +√(v_o^2sin^2θ))

The Attempt at a Solution

We have to calculate the angle
h=(v_o^2sin^2θ)/2g

20.4=25^2sin^2θ/2g

20.4=31.8878sin^2θ

divide both sides by 31.8878

0.639744 = sin^2θ

take the square root of both sides

sinθ = 0.79984

take the inverse sin of both sides

θ = 0.927029

or in degrees

θ = 53.11°

Now that we calculated the angle, we calculate the time to reach the maximum height:

t=v_osinθ/g
=25sin(53.11)/9.8
= 2.04 s


Now we calculate the horizontal displacement at its maximum height

x=v_ocosθt
=25cos53.11*2.04=30.61 m

Now we calculate the range

R=v_ocosθ/g * (v_osinθ +√(v_o^2sin^2θ))
= 25cos53.11/9.8 * (25sin53.11+√(25^2sin^253.11))
= 61.24 m

Horizontal distance from the wall = 61.24-30.61 = 30.63 m

Is this correct? and if it's not tell me where

The Attempt at a Solution

 

[pasmith]

Homework Statement

A catapult is capable of firing a projectile with a velocity of 25 m/s, the catapult is surrounded by a wall of height 20.4 m what is the maximum distance measured from the wall, that the catapult
projectiles are capable of hitting?

I think you also need to know how far the wall is from the catapult - call that x_1. Your approach assumes that the wall is positioned where maximum height is attained (ie, at half the range) and that the trajectory only just clears it, in which case your answer is correct.

For this problem it's probably best to eliminate t and work with
y(x) = x \tan\theta - \frac{gx^2}{2v_0^2 \cos^2\theta}

Since horizontal distance is maximal when \theta = \frac14 \pi, the only question is whether that clears the wall, ie whether
<br /> x_1\tan(\pi/4) - \frac{gx_1^2}{2v_0^2\cos^2(\pi/4)} \geq 20.4.<br />
If not you'll need to take the smallest \pi/4 &lt; \theta &lt; \pi/2 for which the trajectory clears the wall.

Your formula for the range can be simplified: 0 &lt; \theta &lt; \pi/2 so \sin \theta \geq 0 and v_0 &gt; 0. Therefore
R = (v_0/g)\cos\theta (v_0\sin\theta + \sqrt{v_0^2\sin^2\theta})<br /> = (v_0/g)\cos\theta (2v_0\sin\theta) = (v_0^2/g)\sin(2\theta)
The distance beyond the wall where the projectile lands is then (v_0^2/g)\sin(2\theta) - x_1.