Projectile Motion Question

  • Thread starter Jennifer K
  • Start date
  • #1

Homework Statement



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?

Homework 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

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
 

Answers and Replies

  • #2
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 [tex]v^2=u^2-2gh_{max}[/tex] to get the max height.

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

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