# Projectile Motion Firing Question

1. Jun 13, 2010

### iurod

1. The problem statement, all variables and given/known data
A projectile is fired with an initial speed of 65.2m/s at an angle of 34.5deg above the horizontal on a long flat firing range. Determine
(a) the maximum height reached by the projectile,
(b) the total time in the air,
(c) the total horizontal distance covered (that is, the range)
(d) the velocity of the projectile 1.50s after firing

2. Relevant equations
v = vo +at
x = xo + vot + .5at2
v2 = vo2 + 2a(x - xo)
x=vt

3. The attempt at a solution
First I wanted to find out the initial velocity in the x and y direction so i made a right triangle with the hypotenuse as 65.2 m/s with an angle of 34.5 from here I solve for opposite (sin34.5 = opp/65.2) and got 36.929. Then i solved for adjacent (cos34.5 = adj/65.2) and got 53.733. the adjacent is my Xo and the opposite is my Yo.

a. the maximum height reached
Equation v2 = vo2 + 2a(y - yo)
0= 36.9292 + 2(-9.8)(y)
-1363.75 = -19.6y
y = 69.579 i rounded to 69.6m

b. total time in the air
Equation y = yo + vot + .5at2
0 = 0 + 36.929(t)+ .5(-9.8)(t2)
-4.9t2 + 36.929t + 0 = 0
form here I did a quadratic and got t = 7.536 I rounded to 7.5s

c. the total horizontal distance covered (that is, the range)
Equation x = xo + vot + .5at2
x = 0 + 53.733(7.5) +0
x = 402.99 I rounded to 403m

d. the velocity of the projectile 1.50s after firing
Equation vy = voy +at
vy = 36.929 + -9.8(1.5)
vy = 22.229 I rounded to 22.2m/s

Are my solutions and use of formulas for this problem correct?

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

2. Relevant equations

3. The attempt at a solution

2. Jun 13, 2010

### inky

Yes, correct. When you find the time easiest is
v(y)=v(0y)-gt
-v(0)sin(theta)=-v(0)sin(theta)-gt
t=[2v(0)sin(theta)]/g