# Mechanics

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#1
A cannonball is fired horizontally from a seaside tower and hits the sea 2.7s later at a distance of 58m away from the foot of the tower.

What is the horizontal component for velocity

How do you work it out?
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5 years ago
#2
use suvat in the horizontal direction (s=ut+0.5at^2)

so s=58, t=2.7 and a=0

58=2.7u+0.5x0x2.7^2

u=58/2.7=21.48m/s

as a=0, velocity stays at 21.48 m/s
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5 years ago
#3
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5 years ago
#4
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5 years ago
#5
(Original post by 1338)
use suvat in the horizontal direction
We understand you want to help but forum guidelines ask that you do not post the complete solution to a question.
Give a hint as to how to do it.

Secondly, you do not need to use a suvat for the horizontal component of motion as it is uniform velocity.
v = s/t is sufficient.
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#6
(Original post by 1338)
use suvat in the horizontal direction (s=ut+0.5at^2)

so s=58, t=2.7 and a=0

58=2.7u+0.5x0x2.7^2

u=58/2.7=21.48m/s

as a=0, velocity stays at 21.48 m/s
I get confused with the initital and acceleration for vertical and horizontal Could you explain?
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#7
(Original post by Stonebridge)
We understand you want to help but forum guidelines ask that you do not post the complete solution to a question.
Give a hint as to how to do it.

Secondly, you do not need to use a suvat for the horizontal component of motion as it is uniform velocity.
v = s/t is sufficient.
Stonebridge could you explain horizontal and vertical(projectiles) like the acceleration and initial speed i keep getting confused
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5 years ago
#8
If your doing M1, pretty much all you need to know for projectile motion is the acceleration in the vertical direction is (I think always) g which is 9.8ms^-2 towards the ground and acceleration in the horizontal direction is 0. So horizontal velocity doesn't change but vertical velocity does (so you need suvat for those questions).

Hopefully this won't be deemed as too helpful by the mods
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