I need help on projectile motion questions because i had to miss a whole lesson for some reason and my teacher said to try some of the questions.(adsbygoogle = window.adsbygoogle || []).push({});

Anyways im stuck on this question.

A cannon is set at an angle of 45 degrees above the horizontal. A cannon ball leaves the muzzle with a speed of 220 m/s. Air resistance is negligible. Determine the cannonball's

a) maximum height

b) time of flight

c) horizontal range (to the same vertical level)

This is what i did. Im not going to do step by step on the part i got.

Before these i started i found VIx which is 155.6 m/s and Viy is the same

a) At max height Vfy=0( i dont get why but i know thats how you do it)

2ad=Vi^2 + Vf^2

d= (155.6)^2/-2(-9.8)

d= 1240m

b)i have no idea how to get this one i hope you can help me.

c)i started but i couldnt finish. This is what i did

d=Vit + 1/2a(t^2)

1240=155.6t - 4.9t^2

4.9t^2 - 155.6t + 1240 = 0

I know i need to do the quadratic formula after this but i get a negative with the radical so i dont know what to do i think i made a mistake somewhere.

Can you please help me?

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# Homework Help: Need help projectile motion

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