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linear motion help

A ball is thrown vertically up into the air from the edge of a 24.5m high cliff and falls to the base of the cliff. If the initial vertical velocity was 19.6m/s, find the time taken for the ball to land at the foot of the cliff. Answer given is 5 seconds

But I keep getting 2 ot 3 seconds regardless of what SUVAT equation I used. Please could someone help me with this?

Reply 1

username4591252

I got 5 seconds, using s = ut + 1/2 at²

For my SUVAT table, with up as the positive direction:

S: -24.5
U: 19.6
V: x
A: -9.8
T: t

and then solving the quadratic for t

Reply 2

ddsizebra

Original post by FlawlessChicken

I got 5 seconds, using s = ut 1/2 at²

For my SUVAT table, with up as the positive direction:

S: -24.5
U: 19.6
V: x
A: -9.8
T: t

I thought it was s = ut 1/2 at^2

(edited 5 years ago)

Reply 3

username4591252

Yeah I meant s = ut + 1/2 at^2 , it's weird typing maths on here

Original post by ddsizebra

I thought it was s = ut 1/2 at^2

Reply 4

ghostwalker

Original post by FlawlessChicken

Yeah I meant s = ut + 1/2 at^2 , it's weird typing maths on here

Try using LaTeX - userguide here.

\displaystyle s=ut+\frac{1}{2}at^2

Reply 5

ddsizebra

Original post by FlawlessChicken

Yeah I meant s = ut + 1/2 at^2 , it's weird typing maths on here

I've tried that way repeatedly but still getting 2 or 3 not 5.... what I'm I doing wrong?

Reply 6

optalk

Original post by ddsizebra

I've tried that way repeatedly but still getting 2 or 3 not 5.... what I'm I doing wrong?

Well, take 19.6 to be a negative value (it's a vector quantity and thus can have negative values), seeing as it's being thrown upwards. Then all you do is plug in your values.

24.5 = (-19.6)(t) +1/2(9.81)(t)^2

Also, might I add, you'll get two values but since it's asking for time, you take the positive one.