Projectile Motion Bomb Question

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SUMMARY

The discussion focuses on calculating the initial velocity and maximum height of a mortar shell fired at a 45° angle. The shell weighs 120.5 kg and can travel a maximum distance of 5321 m. The initial velocity (u) is determined to be 229 m/s, and the maximum height reached by the shell is calculated to be 1338 m. The range formula R = u² sin(2θ) / g and the maximum height formula max ht = u² sin²(θ) / (2g) are confirmed as essential for these calculations.

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Homework Statement



(b) A power hungry megalomaniac obtains mortar bomb capability. The device could fire 120.5kg mortar shells a maximum distance of 5321m when projected at 45° to the horizontal. What would the velocity of the shell be as it left the muzzle of the device?

(c) What was the maximum height reached by the shell?

Homework Equations



S=Ut + 1/2at^2
v^2=u^2 +2as
v=u+at

where s=distance, u=initial velocity, a= acceleration and t=time

The Attempt at a Solution



t=16.4s
u=229m/s
maximum height=1338m

Can anyone help me check if the solution is right. Thks
 
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just use the range formula , i.e R = u2 sin 2[tex]\theta[/tex] / g

and max ht = u2 sin 2 [tex]\theta[/tex] / 2g
 

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