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Phys student
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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
[tex] h=(v_o^2sin^2θ)/2g[/tex]
[tex] x=v_otcosθ [/tex]
[tex] y=v_otsinθ -0.5gt^2 [/tex]
[tex] R=v_ocosθ/g * (v_osinθ +√(v_o^2sin^2θ)) [/tex]
The Attempt at a Solution
We have to calculate the angle
[tex] h=(v_o^2sin^2θ)/2g[/tex]
[tex] 20.4=25^2sin^2θ/2g [/tex]
[tex] 20.4=31.8878sin^2θ[/tex]
divide both sides by 31.8878
[tex] 0.639744 = sin^2θ [/tex]
take the square root of both sides
[tex] sinθ = 0.79984 [/tex]
take the inverse sin of both sides
[tex] θ = 0.927029 [/tex]
or in degrees
[tex] θ = 53.11° [/tex]
Now that we calculated the angle, we calculate the time to reach the maximum height:
[tex] t=v_osinθ/g[/tex]
[tex] =25sin(53.11)/9.8 [/tex]
[tex] = 2.04 s [/tex]
Now we calculate the horizontal displacement at its maximum height
[tex] x=v_ocosθt [/tex]
[tex] =25cos53.11*2.04=30.61 m[/tex]
Now we calculate the range
[tex] R=v_ocosθ/g * (v_osinθ +√(v_o^2sin^2θ)) [/tex]
[tex] = 25cos53.11/9.8 * (25sin53.11+√(25^2sin^253.11)) [/tex]
[tex] = 61.24 m [/tex]
Horizontal distance from the wall = [tex]61.24-30.61 = 30.63 m[/tex]
Is this correct? and if it's not tell me where