Projectile Motion, Physics I with Calc. I

[bohregard]

Homework Statement

A ball is thrown leftward from the left edge of the roof, at height h above the ground. The ball hits the ground 1.60 seconds later, at distance d = 24.0 m from the building and at angle theta = 55 degrees with the horizontal. (a) Find h. What is the (b) magnitude and (c) angle relative to the horizontal of the velocity at which the ball is thrown?

We know that the impact angle is theta = 55 degrees
x distance is 24.0 m
time for the ball to hit the ground is 1.60 seconds.

Homework Equations

y-y₀ = v₀sin(theta)t - gt²/2

x-x₀ = v₀cos(theta)t

V_x = v_0xt

V_y = v_0yt

v_0x = v₀cos(theta)

v_0y = v₀sin(theta)

The Attempt at a Solution

My first step was to solve for h. What I did was reverse the motion so instead of tossing the ball from the rough, we tossed it from the ground. I made the impact point the origin (0,0). Next I took my second equation and solved it for v₀ which gave me x-x₀ / cos(theta)t = v₀. I plugged that into the first equation and then solved for y which gave me the value 21.73 m.

Now, here is where I get lost. I'm not sure exactly what to solve for. I thought maybe if I solved the equation backwards like I did for height I could find the impact velocity and make that my initial velocity. The problem is I don't know where to start. Any suggestions would be appreciated for sure!

 

[pgardn]

Homework Statement

A ball is thrown leftward from the left edge of the roof, at height h above the ground. The ball hits the ground 1.60 seconds later, at distance d = 24.0 m from the building and at angle theta = 55 degrees with the horizontal. (a) Find h. What is the (b) magnitude and (c) angle relative to the horizontal of the velocity at which the ball is thrown?

We know that the impact angle is theta = 55 degrees
x distance is 24.0 m
time for the ball to hit the ground is 1.60 seconds.

Homework Equations

y-y₀ = v₀sin(theta)t - gt²/2

x-x₀ = v₀cos(theta)t

V_x = v_0xt

V_y = v_0yt

v_0x = v₀cos(theta)

v_0y = v₀sin(theta)

The Attempt at a Solution

My first step was to solve for h. What I did was reverse the motion so instead of tossing the ball from the rough, we tossed it from the ground. I made the impact point the origin (0,0). Next I took my second equation and solved it for v₀ which gave me x-x₀ / cos(theta)t = v₀. I plugged that into the first equation and then solved for y which gave me the value 21.73 m.

Now, here is where I get lost. I'm not sure exactly what to solve for. I thought maybe if I solved the equation backwards like I did for height I could find the impact velocity and make that my initial velocity. The problem is I don't know where to start. Any suggestions would be appreciated for sure!

What is the red equation for?

You sure part b is not missing a word?

 

[bohregard]

What is the red equation for?

You sure part b is not missing a word?

Yup that's the question. The red equation is the y component of velocity.

 

[pgardn]

Yup that's the question. The red equation is the y component of velocity.

Well for projectile motion that will not work as Vy changes through time.

Im thinkin the question intended to ask something about the velocity of the ball when it hits the ground and make some comparison with the initial conditions. But I cannot tell for sure.