- #1
davetheant
- 5
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An arrow is shot at 31.0° above the horizontal. Its velocity is 53 m/s, and it hits the target.
A) What is the maximum height the arrow will attain?
B) The target is at the height from which the arrow was shot. How far away is it?
It says to neglect air resistance so acceleration in the x direction must be 0. Obviously gravity makes the acceleration in the y direction -9.8
I tried to calculate the maximum height by using the formula Max Height = Vi*T+1/2*-9.8*T^2
I got T to equal 2.18 by [53*sin(40)]/9.8
please help
A) What is the maximum height the arrow will attain?
B) The target is at the height from which the arrow was shot. How far away is it?
It says to neglect air resistance so acceleration in the x direction must be 0. Obviously gravity makes the acceleration in the y direction -9.8
I tried to calculate the maximum height by using the formula Max Height = Vi*T+1/2*-9.8*T^2
I got T to equal 2.18 by [53*sin(40)]/9.8
please help