Projectile Motion Firing Question

[iurod]

Homework Statement

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

Homework Equations

v = v_o +at
x = x_o + v_ot + .5at²
v² = v_o² + 2a(x - x_o)
x=vt

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 X_o and the opposite is my Y_o.

a. the maximum height reached
Equation v² = v_o² + 2a(y - y_o)
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 = y_o + v_ot + .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 = x_o + v_ot + .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 v_y = v_oy +at
v_y = 36.929 + -9.8(1.5)
v_y = 22.229 I rounded to 22.2m/s

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

Thanks for your help!

 

[inky]

Homework Statement

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

Homework Equations

v = v_o +at
x = x_o + v_ot + .5at²
v² = v_o² + 2a(x - x_o)
x=vt

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 X_o and the opposite is my Y_o.

a. the maximum height reached
Equation v² = v_o² + 2a(y - y_o)
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 = y_o + v_ot + .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 = x_o + v_ot + .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 v_y = v_oy +at
v_y = 36.929 + -9.8(1.5)
v_y = 22.229 I rounded to 22.2m/s

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

Thanks for your help!

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