Projectile Motion with displacement and time

[sotrashy]

Homework Statement

A boy kicks a ball into the air. It takes 4.3 seconds to land 41.1 meters from his position to the right. What is the launch velocity and launch angle of the ball? Ignore air resistance.

Δx = 41.1 m
t = 4.3 s
a = -9.8 m/s²

Homework Equations

d_x = v_xt
v_fy = v_0y + at

The Attempt at a Solution

d_x = v_xt
d_x / t = v_x
41.1 m / 4.3 s = v_x = 9.56 m/s

v_fy = v_0y + at
0 = v_0y + (-9.8 m/s² * 4.3 s)
0 = v_0y - 42.1 m/s
v_0y = 42.1 m/s

launch velocity = (9.56 m/s)² + (42.1 m/s)² = v_i²
v_i = 43.2 m/s

launch angle = v_xi = v_icosθ
9.56 m/s = 43.2 m/s * cosθ
θ = 77.2°

Can anyone please look through my work to see if I have done it correctly? I'm getting conflicting answers from wolframalpha.

Thanks!

 

[tal444]

Looks all good to me.

 

[sotrashy]

It seemed like that to me, yet when I input these values into http://www.wolframalpha.com/input/?i=projectile+motion, I get travel time that is twice the question's (8.6s) and a horizontal displacement that is also twice the distance (82.2m).

I can't figure out why there's a difference.

 

[gneill]

The Attempt at a Solution

d_x = v_xt
d_x / t = v_x
41.1 m / 4.3 s = v_x = 9.56 m/s

Okay, that looks good.

v_fy = v_0y + at
0 = v_0y + (-9.8 m/s² * 4.3 s)
0 = v_0y - 42.1 m/s
v_0y = 42.1 m/s

Oops. The vertical velocity won't be zero when it lands --- it's going to impact the ground at some speed (and probably bounce). (When is the vertical velocity zero?)

You can use this kinematic relationship if you choose the appropriate time when the velocity is zero, otherwise find another equation for the vertical vertical trajectory that includes the landing condition (Hint: height at landing is zero).

 

[tal444]

Whoops, made a mistake in my calculations. You should probably try d = vt + 1/2at² instead. As gneill just said, your V_fy is 0 at the TOP of it's trajectory, not the bottom.

 

[sotrashy]

I used the calculator at http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html#tracon while inputting Vxi = 9.56, t = 4.3s, and dx = 41.1 and I get the vertical initial velocity to be 42.1 m/s which is what I got in the beginning.

I still don't know where I'm going wrong :S

 

[sotrashy]

Ahhh, I see. I'm calculating AT time 4.3s!

dy = 0 = vt + 1/2at^2

Better?

 

[tal444]

yesh