Projectiles Launced at an Angle

[Tnn Ace03]

I have a test coming up on wednesday and i have a question about projectiles launced at an angle. Our teacher tells us that we can substitute equation into anontehr to get the final answer

A quarterback throws the football to a reciever who is 31.5 meters down the field. If the football is thrown at an initial angel of 40.0 degrees to the horizontal, at which initail spped must the quarterbak throw the ball? what is the ball's highest point during the flight.


The answer is

17.7 m/s as initail velocity

Delta Y= 6.60 meters

My question is how do you get the final answer and put it into simple terms

 

[radou]

I suggest you make a search, since projectile motion is a frequent homework question. I believe it will be enough just to type 'projectile motion' into the search box. Read some posts, and everything should be more clear. If not, we'll try to make it clear. :)

 

[kyle8921]

To get the final answer, you can use the equations of projectile motion, specifically the equations for displacement (Δy) and initial velocity (v0) in the vertical direction. These equations are:

Δy = v0sinθt + (1/2)gt^2

and

v0sinθ = gt

where θ is the initial angle, t is the time of flight, and g is the acceleration due to gravity (9.8 m/s^2).

To find the initial velocity, you can use the second equation and solve for v0:

v0 = gt/sinθ

Substituting in the given values of θ = 40 degrees and Δy = 31.5 meters, we get:

v0 = (9.8 m/s^2)(31.5 m)/(sin 40°) ≈ 17.7 m/s

This is the initial speed at which the quarterback must throw the ball.

To find the ball's highest point during flight, we can use the first equation and solve for t (the time it takes for the ball to reach its highest point):

Δy = v0sinθt + (1/2)gt^2

Rearranging and setting Δy = 0 (since the ball's highest point is when it reaches the same height as where it was thrown from), we get:

0 = v0sinθt + (1/2)gt^2

Solving for t, we get:

t = 2v0sinθ/g

Substituting in the values of v0 and θ, we get:

t = (2)(17.7 m/s)(sin 40°)/(9.8 m/s^2) ≈ 2.08 seconds

To find the ball's highest point, we can plug this value of t into the first equation:

Δy = (17.7 m/s)(sin 40°)(2.08 s) + (1/2)(9.8 m/s^2)(2.08 s)^2 ≈ 6.60 meters

This is the ball's highest point during flight.

In simple terms, to find the initial speed, you need to use the equation v0 = gt/sinθ, where g is the acceleration due to gravity and θ is the initial angle. To find the ball's highest point, you need