# Very easy kinematics question

#### iamaelephant

##### Diamond Member
I'm going through some 1-dimensional kinematics problems and found this in my text book:

A projectile is launched straight up at 60.0 m/s from a height of 80.0 m, at the edge of a sheer cliff. The projectile falls, just missing the cliff and hitting the ground below. Find (a) the maximum height of the projectile above the point of firing, (b) the time it takes to hit the ground at the base of the cliff, and (c) its velocity at impact.

(a) 184 m Correct
(b) 13.5 s Correct
(c) -112.3 m/s Wrong

I got that answer using the equation:
v = v0 + at
= 20.0 + (-9.8)(13.5)
=-112.3

The answer in the text book is -72.3 m/s so what did I do wrong?

#### Eeezee

##### Diamond Member
Originally posted by: iamaelephant
Christ you're right.... where the hell did i get 20m/s bed time methinks.

Edit - time value was right, just v0 was wrong.

I can see where you may have made the mistake. You subtracted 60 from 80 to get 20. Doing introductory physics this late can screw with your head.

A good thing to remember is be VERY mindful of your units. You must never subtract two quantities with different units; you can multiply stuff together as much as you want to get the units you want (m/s * s * m * kg * C = kgCm^2). You can't add a meters/second quantity to a meters quantity, it just doesn't make sense! You're probably aware of this already so I'm going to stop talking now.

#### msparish

##### Senior member
v = v0 + at

v0 = 60m/s, not 20m/s.

#### KillerCharlie

##### Diamond Member
and you're using the wrong value for time

#### iamaelephant

##### Diamond Member
Christ you're right.... where the hell did i get 20m/s bed time methinks.

Edit - time value was right, just v0 was wrong.

#### Goosemaster

##### Lifer
Originally posted by: iamaelephant
Christ you're right.... where the hell did i get 20m/s bed time methinks.

Edit - time value was right, just v0 was wrong.

lol..that's what happens when you physics at this hour

<---does it all during hte week

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