Projectile Motion Firing Question

In summary: a. v(y)=v(0y)-gtvy=[v(0)sin(theta)]-gt0=[v(0)sin(theta)]-gtt=[v(0)sin(theta)]/gb. t=[2v(0)sin(theta)]/gt=[2(65.2)sin(34.5)]/9.8t=7.51 secondsc. x=v(0x)tx=(v(0)cos(theta))(7.51)x=403 metersd. v(x)=v(0x)v(x)=v(0)cos(theta)v(x)=53.7m/s
  • #1
iurod
51
0

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 = vo +at
x = xo + vot + .5at2
v2 = vo2 + 2a(x - xo)
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 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?

Thanks for your help!
 
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  • #2
iurod said:

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 = vo +at
x = xo + vot + .5at2
v2 = vo2 + 2a(x - xo)
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 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?

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
 

1. How does the initial velocity affect the trajectory of a projectile?

The initial velocity of a projectile, which is the speed and direction at which it is launched, plays a significant role in determining its trajectory. The higher the initial velocity, the farther and higher the projectile will travel. The angle at which the projectile is launched also affects its trajectory, as a higher launch angle will result in a longer flight time and a steeper curved path.

2. What is the formula for calculating the maximum height of a projectile?

The formula for calculating the maximum height of a projectile is h = (v02sin2θ)/2g, where h is the maximum height, v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity. This formula assumes a level launch and landing point and does not take into account air resistance.

3. How does air resistance affect the trajectory of a projectile?

Air resistance, also known as drag, can significantly impact the trajectory of a projectile. As the projectile travels through the air, it experiences resistance due to the air molecules it comes in contact with. This resistance can slow down the projectile and alter its trajectory, causing it to fall short of its expected landing point. This is why projectiles with the same initial velocity but different shapes (such as a bullet and a feather) will follow different paths due to differences in air resistance.

4. What is the difference between horizontal and vertical motion in projectile motion?

In projectile motion, there are two components of motion: horizontal and vertical. Horizontal motion is the motion of the projectile parallel to the ground, while vertical motion is the motion perpendicular to the ground. In the absence of external forces, horizontal motion is constant and unaffected by gravity, while vertical motion is affected by gravity and follows a parabolic path.

5. Can the range of a projectile be increased by changing the launch angle?

Yes, the range of a projectile can be increased by changing the launch angle. The optimal launch angle for maximum range is 45 degrees, as this angle will give the projectile the longest flight time and the farthest horizontal distance. However, the range can also be affected by the initial velocity and air resistance, so changing the launch angle may not always result in a significant increase in range.

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