Projectile Motion Question: A projectile is fired on a long flat firing range....

[blackcemre]

Homework Statement

A projectile is fired with an initial speed of 46.6 m/s at an angle of 42.2 ∘ above the horizontal on a long flat firing range.

1-Determine the maximum height reached by the projectile.

2-Determine the total time in the air.

3-Determine the total horizontal distance covered (that is, the range).

4-Determine the speed of the projectile 1.50 s after firing.

5-Determine the direction of the motion of the projectile 1.50 s after firing.

Homework Equations

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

The Attempt at a Solution

A.) Determine the maximum height Ymax reached by the projectile.

Ymax = (V*sin 45.2)^2/2g = (43.6*0.71)^2/19.612 = 48.86 mB.)Determine the total time t in the air.

t = 2*V*sin 45.2/g = 2*43.6*0.71/9.806 = 6.314 secC.)Determine the total horizontal distance covered (that is, the range).

Δx = V*cos 45.2*t = 43.6*0.705*6.314 = 194.08 m

 

[ZapperZ]

Homework Statement

A projectile is fired with an initial speed of 46.6 m/s at an angle of 42.2 ∘ above the horizontal on a long flat firing range.

1-Determine the maximum height reached by the projectile.

2-Determine the total time in the air.

3-Determine the total horizontal distance covered (that is, the range).

4-Determine the speed of the projectile 1.50 s after firing.

5-Determine the direction of the motion of the projectile 1.50 s after firing.

Homework Equations

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

The Attempt at a Solution

A.) Determine the maximum height Ymax reached by the projectile.

Ymax = (V*sin 45.2)^2/2g = (43.6*0.71)^2/19.612 = 48.86 mB.)Determine the total time t in the air.

t = 2*V*sin 45.2/g = 2*43.6*0.71/9.806 = 6.314 secC.)Determine the total horizontal distance covered (that is, the range).

Δx = V*cos 45.2*t = 43.6*0.705*6.314 = 194.08 m

So what's the question? You couldn't do #4 and #5?

Zz.

 

[gneill]

A projectile is fired with an initial speed of 46.6 m/s at an angle of 42.2 ∘ above the horizontal

A.) Determine the maximum height Ymax reached by the projectile.

Ymax = (V*sin 45.2)^2/2g = (43.6*0.71)^2/19.612 = 48.86 m

The numbers you've used in your work do not match those stated in the problem statement. Which numbers are correct?

 

[Chandra Prayaga]

There is confusion about the symbols used.
Is V the initial speed?
In the relevan equations, there are two different equations for x.