A ball is shot at an angle of 45 degrees into the air with initial velocity of 40 ft/sec.
a) Assuming no air resistance, how high does it go?
b) How far away does it land?
im not sure how to do this problem, the examples from the book doesnt help.
A ball is shot at an angle of 45 degrees into the air with initial velocity of 40 ft/sec.
a) Assuming no air resistance, how high does it go?
b) How far away does it land?
im not sure how to do this problem, the examples from the book doesnt help.
the ball is a projectile in a conservative field with force only in one direction
just apply equations of motion in horizontal and vertical direction
to get h=(v^2)* sin^2( x)/2g
x is angle of projection and v is initial velocity
this is quite standard and does not come under calculus
I'm not objecting to srinivas' answer, I just think it's good to see how to do this from the equations of motion.
So, I am picking the "usual" coordinate system, with +x to the right and +y upward. My origin is at the point where the ball is being shot at, which is (presumably) at ground level. I am shooting the ball at an angle of 45 degrees above the + x axis with an initial speed of 40 ft/s. The ball (presumably) lands on the ground at the same height it started.
We have the following information:
$\displaystyle \theta = 45^o$
$\displaystyle v_0 = 40$ ft/s
$\displaystyle x_0 = 0$ ft
$\displaystyle y_0 = 0$ ft
$\displaystyle v_{0x} = 40 \cdot cos 45^o$ ft/s
$\displaystyle v_{0y} = 40 \cdot sin 45^o$ ft/s
$\displaystyle a_x = 0$ ft/s^2
$\displaystyle a_y = -32$ ft/s^2
We wish to find first the maximum height. This is the point in the flight where the vertical component of the velocity is 0 ft/s, so we are looking for the point where $\displaystyle v_y = 0$.
$\displaystyle v_y^2 = v_{0y}^2 + 2a_y(y - y_0)$
or
$\displaystyle 0 = v_{0y}^2 + 2a_yy$ in this case.
$\displaystyle y = -\frac{v_{0y}^2}{2a_y} = -\frac{(40 \cdot sin45^o)^2}{-2 \cdot 32} = 12.5$ ft
The second problem is to find the horizontal range. This is the x position when y = 0 ft. (t > 0 s).
So
$\displaystyle x = x_0 + v_{0x}t + (1/2)a_xt^2$ is the only equation of use when the acceleration in the x direction is 0.
$\displaystyle x = v_{0x}t$ in this case.
We need to find out how long the ball is in the air. So when is y = 0 ft?
So
$\displaystyle y = y_0 + v_{0y}t + (1/2)a_yt^2$
$\displaystyle 0 = v_{0y}t + (1/2)a_yt^2$ in this case.
$\displaystyle t = 0$ s or $\displaystyle 0 = v_{0y} + (1/2)a_yt$
The first solution is for when the ball is being shot, so we don't care about it. So we solve the second equation:
$\displaystyle 0 = v_{0y} + (1/2)a_yt$
$\displaystyle t = - \frac{2v_{0y}}{a_y} = - \frac{2 \cdot 40 \cdot sin 45^o}{-32} = 1.76777$ s
Then
$\displaystyle x = v_{0x}t = 40 \cdot cos 45^o \cdot 1.76777 = 50$ ft. (I've kept a few extra digits on the time that I didn't show above during this calculation.)
-Dan
I managed an A+ and a B+ on the last two, so I don't think I'm going to take it. I'm worried about the final, but I'll take what I get at this point. I'm looking forward to next semester's teacher being a little easier to work with. For me, this is only the beginning, as I am in a heavy math intensive science major. I never took pre-calc or trig, so I've been playing quite a bit of catch up.
I don't even know where to start with half of these problems.