Projectile Motion Question

1. Mar 14, 2007

Jennifer K

1. The problem statement, all variables and given/known data

Hello i'd apprcieate it if someone could help me with this projectile motion problem.

A projectile off mass 33.6 grams is thrown from a hight of 1 meter of the ground at an angle of 60 degress, it lands 10 meters away from the point it was thrown.

What is the peak vertical distance?
What is Vi?

2. Relevant equations

(1) V = U+at^2

(2) s = 1/2 (U+V)t

(3) s = Ut + 1/2 at^2

(4) V^2 = U^2 +2as

3. The attempt at a solution

I tried to do it in two parts:

(1)

U = ViSIN60

V = 0

a = -9.81 m/s^2

s1 = ?

t = ?

(2)

U = 0

V = ?

a = -9.81

Stotal = S1 + 1 meter

t2 = ?

and then for the horizontal:

U = ViCOS60

V = ?

a = 0

S = 10 meters

ttotal = t1 + t2

i'm really stuck if anyone can give me any tips i will really appreciate it. thank youuuuuu :) :) :)

Jenny

2. Mar 15, 2007

chaoseverlasting

The time it took to travel 10 meters is t=10/vcos60. This is the same time it took to reach the max height, h, and fall h+1 meters.
Individually find the time it took the projectile to reach the max height, and then fall h+1 meters. This time is equal to t.
You can use $$v^2=u^2-2gh_{max}$$ to get the max height.

This should give you enough equations to solve for the variables.