An arrow is shot at 28.0° above the horizontal. its initial speed is 50 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?

 We will need to work with the components of the velocity, in the x and the y direction. We will say up is positive so g is -9.81 m/s^2. 

Given that the angle was 32 degrees: 

Velocity up (in the y direction) is 55 m/s * sin 32 = 29.15 m/s 
And 

Velocity forward (in the x direction) is 55 m/s * cos 32 = 46.64 m/s 

The acceleration of gravity, -9.81 m/s2 continuously decreases the velocity in the y direction. At the maximum height, the velocity will be zero. This should make sense, for as soon as the decreasing velocity becomes negative, the arrow will start to fall. 

We have v = v(0) + at 

And we set this to zero and solve for t: 

0 = 29.15 + -9.81t 

9.81t = 29.15 

t = 2.97 seconds 

To calculate height at this point, we use the equation that calculates position based on time, acceleration, and initial velocity (we could use an alternate too, an equation derived from the one we are now using and v = v(0) + at. 

x = x(0) + v(0)t + (1/2)at^2 

x = 0 + 29.15 * 2.97 + 0.5 9.81 (2.97)^2 

x = 43.30 m 

For a projectile, the plot of distance traveled in the upward direction is a parabola, and it takes the same amount of time to come down as it did to go up. 

We can double 2.97 to get the time of impact on the target at 2(2.97) = 5.94 seconds 

(Alternately, if you like, you can solve 

0 = 0 + 29.15t + 0.5 9.81 t^2 

And find that the two roots are 0 and 5.94). 

http://www.math.com/students/calculators... will do the quadratic for you. 

Given a horizontal velocity of 46.64 m/s, we can calculate 

46.64 m/s (5.94 s) = 277 m for the distance of the target.