Question

Please help me with 2 brief questions.

Answer 1

A cannonball is shot (from ground level) with an initial horizontal velocity of 33 m/s and an initial vertical velocity of 25 m/s.
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\(\rm \large I~found~so~far:\)

[1] initial speed, \(\rm u=41.4(m/s)\)
[2] angle of cannonball with respect to ground, \(\rm \theta=37.15º\)
[3] maximum height the cannonball goes above the ground \(\rm x=31.8(m)\)
[4] horizontal displacement is \(\rm 167.6(m)\)
[5] the entire time of the ball motion \(\rm 5.1(s)\)

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\(\rm \large I~need:\)

[6] speed of the cannonball at, \(\rm t=3.7(s)\)
[7] height (Vertical Displacement) at, \(\rm t=3.7(s)\)

Answer 2

6.Vi = √[Vx²+Vy²] = 44.2 m/s
7.3) Hmax = Vi²sin²Θ/(2g) = 62.49 m

Answer 3

Are you sure? Becsause that doesn't seem to be right.

Answer 4

Don't know,I never liked physics. Maybe he can help @Hero @dan815 @iambatman

Answer 5

@triciaal

Answer 6

I found the vertical displacement x=25.35m
applying: \(x = ut + \dfrac{1}{2}at^2\)

I have put:
t=3.7s
a=-9.81m/s\(^2\)
u=25m/s

Answer 7

Nice.

Answer 8

I need velocity at t=3.7

the rest i found. tnx for coming btw

Answer 9

Your welcome!

Answer 10

A cannonball is shot (from ground level) with an initial horizontal velocity of 33 m/s and an initial vertical velocity of 25 m/s.

[6] speed of the cannonball at, t=3.7(s)
[7] height (Vertical Displacement) at, t=3.7(s)

Answer 11

[7]=25.35m found that:)