Help w/ projectiles??!!(the last one on my hw that i cant solve!!!!)
Favorites|Homepage
Subscriptions | sitemap
HOME > Mathematics > Help w/ projectiles??!!(the last one on my hw that i cant solve!!!!)

Help w/ projectiles??!!(the last one on my hw that i cant solve!!!!)

[From: ] [author: ] [Date: 11-10-19] [Hit: ]
47 seconds for time but i dont think thats right. when i use 1.47 for time i keep getting 10.62meters for the answer and my teacher said its wrong. >.-Answer should be around 65.......
An arrow is shot at an angle of 33º above the horizontal with a velocity of 26.5 m/s. When the arrow lands back on the ground, how far did it travel horizontally? Round your answer to 2 decimal places.


i got 1.47 seconds for time but i dont think thats right. when i use 1.47 for time i keep getting 10.62meters for the answer and my teacher said its wrong. >.<

-
Answer should be around 65.5 meters.

Looks like you might have solved for t at the top of the arrow's trajectory instead of when it hits the ground. It should actually be twice that (t ≈ 2.95 s). Here's what you do. Set up the equation for vertical motion (in the y direction):

d(t) = (1/2)at² + v(initial)t + d(initial)

where:

a = -9.8 m/s²
v(initial) = 26.5*sin(33) m/s
d(initial) = 0 m

Now, you want to solve for when the position (height) equals 0. There should be two times this happens, at t = 0 (since that's your initial position) and when it hits the ground.

(1/2)at² + v(initial)t + d(initial) = 0

-4.9t² + 26.5*sin(33)t = 0

t (-4.9t + 26.5*sin(33)) = 0

Taking out the t = 0 part, which we already know as the initial position, the equation will also be zero when:

-4.9t + 26.5*sin(33) = 0

-4.9t = -26.5*sin(33)

t = (26.5*sin(33)) / (4.9) ≈ 2.95 s

So, now you set up the equation for horizontal motion (in the x direction):

d(t) = (1/2)at² + v(initial)t + d(initial)

where:
a = 0 m/s²
v(initial) = 26.5*cos(33)
d(initial) = 0 m
t = (26.5*sin(33)) / (4.9) ≈ 2.95

d(t) = 26.5*cos(33)*(t)

If you use the exact value for t of (26.5*sin(33)) / (4.9) seconds, then you get a distance of 65.46 meters. If you use the rounded figure of 2.95 seconds, then you get a distance of 65.56 meters. Just depends on how exact your teacher wants you to be, and when you are supposed to round.

Hope that helps!
1
keywords: that,one,solve,my,cant,Help,last,projectiles,the,hw,on,Help w/ projectiles??!!(the last one on my hw that i cant solve!!!!)
New
Hot
© 2008-2010 http://www.science-mathematics.com . Program by zplan cms. Theme by wukong .