# A ball is thrown up at 20 m/s at an agle of 15 degrees

• WeldingStuff
In summary: Therefore, the total time it takes for the ball to hit the ground is twice the time it took to get to the top. In summary, the ball will take approximately 2.2 seconds to hit the ground.
WeldingStuff

## Homework Statement

A ball is thrown upwards at 20 m/s at an angle of 15 degrees to ground.
How long until the ball touches the ground?

Givens for vertical:
Acceleration = -9.8m/s2
v1= 5.2 m/s (20SIN15)
v2=0
displacement=0
t=?

Givens for horizontal
Acceleration= 19.3 m/s (20COS15)
Displacement= ? (perhaps work back from displacement on horizontal to get the time?)
t=?

## Homework Equations

DeltaD= (V1)(DeltaT) + (.5)(A)(DeltaT)
Answer should be 0 OR 1.1s for T ( I don't understand how you are to get two answers)

## The Attempt at a Solution

0=(5.2)(detlaT) +(0.5)(-9.8)(deltaT)[/B]
?? ?? ??
The textbook suggests to use this equation and not the other 4.

Also I am not sure if displacement is zero for the horzontal given that V=D/TStumped for hours.
Nobody I can ask for help

To get the answer the problem requires, I suggest that you just focus on the vertical velocity, with the gravitational acceleration as a constant.

I get that. I don't understand why the book suggests there are two answers 0 and 1.1
I am missing something basic I think.

That's because at the first ##(t=0)## the ball is also on the ground, so if you use the displacement to solve the quadratic equation, you will get two answer, of course one of which is trivial.

Yeah, it's an issue with my math. I have not taken math in high school since the original September 11th, so you can guess my age...
What is the operation called apart from quadratic equation. I have to figure this stuff out from square one. Thank you for the help.

If you set the time ##t## and the displacement ##d,## you can get
$$v_{0,ver}t+\frac{1}{2}(-g)t^2=0$$
and one of the solutions is ##0.##
The way I suggested above:
$$-v_{0,ver}=v_{0,ver}+(-g)t,$$
which can be intuitively get by
$$t=\frac{2v_{0,ver}}{g}$$
by the process of constant acceleration.

When solving projectile motion problems, it tends to be in your best interest to stick with the variable you have the most information on. Projectile motion problems tend to require you to first find t, which is the only variable shared by both x and y, as both are functions of t. If you, for instance, were asked to find the displacement in x instead of finding t, then you would have had to find t first and then use kinematics equations on the x plane instead of the y plane.

WeldingStuff said:

## Homework Statement

A ball is thrown upwards at 20 m/s at an angle of 15 degrees to ground.
How long until the ball touches the ground?

Givens for vertical:
Acceleration = -9.8m/s2
v1= 5.2 m/s (20SIN15)
v2=0
displacement=0
t=?

Givens for horizontal
Acceleration= 19.3 m/s (20COS15)
Displacement= ? (perhaps work back from displacement on horizontal to get the time?)
t=?

## Homework Equations

DeltaD= (V1)(DeltaT) + (.5)(A)(DeltaT)
Answer should be 0 OR 1.1s for T ( I don't understand how you are to get two answers)

## The Attempt at a Solution

0=(5.2)(detlaT) +(0.5)(-9.8)(deltaT)[/B]
?? ?? ??
The textbook suggests to use this equation and not the other 4.

Also I am not sure if displacement is zero for the horzontal given that V=D/TStumped for hours.
Nobody I can ask for help
A simple way to solve this problem is to find the initial vertical velocity of the ball. When the ball is at the top of its arc, its velocity is zero. Using kinematic equations, the time it took for it to reach the top can be found. Due to the conservation of energy, the time it took to get to the top will be equal to the time it takes to hit the ground.

## 1. What is the initial velocity of the ball?

The initial velocity of the ball is 20 m/s. This means that when the ball is first thrown, it is moving at a speed of 20 meters per second.

## 2. What does the angle of 15 degrees represent?

The angle of 15 degrees represents the direction in which the ball is thrown. In this case, the ball is thrown at an upward angle of 15 degrees.

## 3. How high will the ball go?

Using the equations of motion and assuming no air resistance, the ball will reach a maximum height of approximately 3.06 meters.

## 4. How long will the ball stay in the air?

Again, using the equations of motion and assuming no air resistance, the ball will stay in the air for approximately 1.42 seconds before it falls back to the ground.

## 5. What is the velocity of the ball when it reaches its maximum height?

The velocity of the ball when it reaches its maximum height will be 0 m/s. This is because at the highest point of its trajectory, the ball momentarily stops moving upwards and begins to fall back down due to the force of gravity.

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