Two-dimensional kinematics: Projectile Motion

[macha9907]

Homework Statement

A shot-putter throws the shot with an initial speed of 11.2 m/s from a height of 5.00 ft above the ground. What is the range of the shot if the launch angle is a) 20 degrees, b) 30 degrees, or c) 40 degrees?

Homework Equations

y = 1/2 gt^2
x = vt
R= Range
R = (v0^2/g)sin2(theta)
?

The Attempt at a Solution

For a) 20 degrees:
t = sqrt(2y/g)
t = 1.01s
x = (11.2 m/s)(1.01s) = 11.3 m

b) R= [(11.2m/s)^2 /9.81 m/s^2] (2sin(30))
R= 12.78 m (this is incorrect..)

c) R= [(11.2 m/s)^2 /9.81 m/s^2] (2sin(40))
R = 16.48 m (this is also incorrect)

I don't really understand how I should approach this problem, or which equations I should use. The range equation is not working for me. I can't repeat the steps from part a because they did not include the angle in the first place..

I appreciate any help given.

 

[unrulypanda]

For b) Find the time. There are two parts because one half of the parabola has a different floor height.

v[SUB]initialx[/SUB]=v[SUB]initial[/SUB]sin(θ)
v[SUB]final[/SUB]=v[SUB]initial[/SUB]+at
t[SUB]up[/SUB]=-v[SUB]initial[/SUB]/g

v[SUB]final[/SUB][SUP]2[/SUP]=v[SUB]initial[/SUB][SUP]2[/SUP]+2ad
(d+5)=1/2at2
t[SUB]down[/SUB]=√[2(d+5)/g]

t[SUB]up[/SUB]+t[SUB]down[/SUB]=t[SUB]total[/SUB]

The horizontal component maintains a constant velocity until it hits the floor.

v[SUB]y[/SUB]=v[SUB]initial[/SUB]cos(θ)
v[SUB]y[/SUB]=R/t[SUB]total[/SUB]
R=v[SUB]y[/SUB]t[SUB]total[/SUB]

I'm just a student myself so please tell me if you spot anything wrong! The steps above should apply similarly to question C as well.

 

[dacruick]

yes your problem is that you aren't separating components of velocity. You can use the vertical component to calculate the time in the air. and you can use the horizontal component and the time to find the range.