# Velocity and height of a ball thrown at an angle

• enantiomer1
In summary: Convert 390 feet into meters first, then plug that value into your equation.In summary, the initial speed of the ball was 19.45 m/s and the ball rose approximately 37.5 feet above the point where it struck the bat.
enantiomer1

## Homework Statement

A batted baseball leaves the bat at an angle of 35.0 degrees celcius above the horizontal and is caught by an outfielder 390 ft from home plate at the same height from which it left the bat. what was the initial speed of the ball and How high does the ball rise above the point where it struck the bat?

## Homework Equations

D= v0t
D= v0sin 35 degrees

## The Attempt at a Solution

I tried to figure it out using both the equations, but I guess I must be doing something wrong, does anyone know what equation I'm not using?

First, angles do not have units in degrees celsius; angles can be in degrees or radians. You probably had a mind slip, right? =)

Second, the equations that you posted aren't useful right now, but they could be later depending on your problem-solving approach.

Let's work on the first listed unknown, the initial speed of the ball. What is the given data? We know the given angle (35 degrees), and we know that total horizontal distance the ball travels (390 feet--don't forget to convert to meters, if the answer requires it). The ball initially leaves the bat at some height that is unknown, but we do know that the ball is caught at exactly this same height. There is a specialized term to describe such a horizontal distance. What is this term? Do you have any equations for such motion that use this term?

The equation I was looking for is v0= sqrt(R[which in this case is just x]*g/sin(2 theta).
However everytime I plug it in I get sqrt(390*9.81/sin(70))= 63.81 ft/s, converted to m/s this is 63.81 ft/s * .3048 = 19.45 and this is STILL wrong... what am I not adding?

Yes, as you have demonstrated, you want to use the range equation here.

Your problem is that you did not covert 390 ft into meters first. Notice that you have 390 * 9.81; 390 is in units of feet and 9.81 is in units of "meters" per second-squared. Your mixing the length units.

## What is the formula for calculating the velocity of a ball thrown at an angle?

The formula for calculating the velocity of a ball thrown at an angle is v = u + gt, where v is the final velocity, u is the initial velocity, g is the acceleration due to gravity (typically 9.8 m/s^2), and t is time.

## How does the angle affect the velocity of a thrown ball?

The angle at which a ball is thrown affects its velocity because it determines the initial vertical and horizontal components of the velocity. A ball thrown at a higher angle will have a greater vertical velocity and a smaller horizontal velocity compared to a ball thrown at a lower angle.

## What is the relationship between velocity and height of a thrown ball?

The relationship between velocity and height of a thrown ball is that the velocity at any point is equal to the square root of the initial velocity squared plus two times the acceleration due to gravity multiplied by the change in height. In other words, the higher the velocity, the higher the ball will go.

## What is the maximum height a ball can reach when thrown at an angle?

The maximum height a ball can reach when thrown at an angle is determined by the initial velocity and the angle of the throw. The formula for maximum height is h = (u^2 * sin^2θ) / 2g, where h is the maximum height, u is the initial velocity, θ is the angle of the throw, and g is the acceleration due to gravity.

## How can the velocity and height of a thrown ball be calculated in real-life scenarios?

In real-life scenarios, the velocity and height of a thrown ball can be calculated by measuring the initial velocity and angle of the throw, and using the equations mentioned above. However, factors such as air resistance and variations in acceleration due to gravity may affect the accuracy of the calculated values.

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