How do you find velocity given angle and highest point of object?

Answer 1

Use the equation square root of (gravity times distance)/(2 sin theta*cos theta) when the height difference between the initial and final is negligible, meaning the same. If different heights, use the same without the 2 on the bottom. Use the equation square root of (gravity times distance)/(2 sin theta*cos theta) when the height difference between the initial and final is negligible, meaning the same. If different heights, use the same without the 2 on the bottom.

Answer 2

If this is a problem associated with gravity, then gravity will accelerate an object toward earth. And, because it is at some height at the beginning of its fall, we can make some calculations to see how fast it is going at any given moment (air resistance aside) and also how far it has fallen. There is a gravitational force associated with the earth. It pulls on stuff. And that force is fairly constant from place to place. We assign it a fixed value called the gravimetric constant (g0, gn or sometimes just g), and it is said to be 9.80665 m/s2 or about 32.174 ft/s2 as agreed upon by international concensus. We have to have something to work with because of the slight variations in gravity at different places around the globe. And g is the value we go with. Think of it as an "average" if you want to. With the gravitational constant, we can determine that the velocity at the end of one second of fall will be about 32.174 feet per second. After two seconds, it is moving 64.248 feet per second. But that's finding velocity given time. If we know how high the object was at the beginning of its fall, we can, with the gravitational constant, calculate how fast it will be going after any distance it has fallen. There is a formula for that, of course. vfinal2 = vinitial2 + [2g (sfinal - sinitial)] The square of the final velocity equals the square of the initial velocity plus the product of twice the acceleration due to gravity times the difference in the altitudes or heights of the falling body from beginning to end.

Answer 3

How do you find velocity given angle and highest point of object?

Let's say this is a problem about a ball shot out of a cannon. When the cannon is fired, the ball has 2 components of its velocity, vertical and horizontal.

The initial vertical velocity = V * sin θ , Vi = V * sin θ

When the ball reaches the top of its path it stops going up, which means the vertical velocity = 0 m/s.

The final vertical velocity for the up trip = 0 m/s,Vf = 0

.

Average velocity = (vi + vf) ÷ 2

.

Average velocity = (V * sin θ + 0) ÷ 2

.

Eq.#1..Average velocity = (V * sin θ) ÷ 2

.

Due to gravity, the ball's vertical velocity decreases by 9.8 m/s each second.

Vi = V * sin θ

Vf = 0

Change in velocity = Vfinal - Vinitial

Acceleration due to gravity = -9.8 m/s^2

Vfinal - Vinitial = acceleration * time

(0 - V * sin θ) ÷ -9.8 = time

.

Time = -V * sin θ ÷ -9.8

.

Eq.#2..Time = V * sin θ ÷ 9.8

.

Distance = Average velocity * time

The distance upward is Height.

Eq.#3..Height = Average velocity * time

Substitute Eq.#1 for average velocity and Eq.#2 for Time into Eq.#3.

Height = [(V * sin θ) ÷ 2]* [V * sin θ ÷ 9.8]

Eq.#4.. Height = (V^2 * sin^2 θ) ÷ (2 * 9.8)

The general equation would be

Eq.#5.. Height = (V^2 * sin^2 θ) ÷ (2 * g), g = acceleration due to gravity.

Solve Eq.#5 for velocity;… multiply both sides by 2g

Height * 2g = (V^2 * sin^2 θ)…divide both sides by sin^2 θ

(Height * 2g) ÷ sin^2 θ = V^2

V^2 = (Height * 2g) ÷ sin^2 θ

Eq.#6…V = [(Height * 2g) ÷ sin^2 θ]^0.5

Now if we had values of height and angle, we could find V

Answer 4

The formula for finding that out is velocity = distance / mass