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. >.<
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. >.<
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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!
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!