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

## 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 m

B.)Determine the total time t in the air.

t = 2*V*sin 45.2/g = 2*43.6*0.71/9.806 = 6.314 sec

C.)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

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## Answers and Replies

ZapperZ
Staff Emeritus
Science Advisor
Education Advisor

## 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 m

B.)Determine the total time t in the air.

t = 2*V*sin 45.2/g = 2*43.6*0.71/9.806 = 6.314 sec

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