The quarterback of a football team gets off a perfect pass at an angle of 30 degrees with the horizontal. His receiver gathers it in 12.2m downfield from the passer at the same height above ground as the passer's throwing hand.
a. What was the ball's velocity as it left the passer?
b. How high above the passer's outstretched arm did it rise?
For part a. I got a answer of 11.7m/s by using the substitution method. Is this right?
I'm having a bit of trouble finding part b. I'm using the equation s=Vi*t+1/2a*t^2
V = Vi up and u horizontal
u = V cos 30 = .866 V
Vi = V sin 30 = .5 V
in air t seconds
horizontal problem:
12.2 = u t
12.2 = .866 V t
V t = 14.1
so 2Vi t = 14.1 and Vi t = 7.05
vertical problem
h = Vi t - 4.9 t^2 = 0 at start and finish
0 = 7.05 - 4.9 t^2
t = 1.2 seconds in air
V =14.1/t = 14.1/1.2 = 11.75
agree with you
max height at t/2 or at 0.6 seconds
h = 3.5 - 4.9(.36)
about 1.75 meters
To solve part a, you can use the kinematic equation for horizontal motion:
s = v*t
where s is the horizontal distance (12.2m), v is the velocity, and t is the time (which will be the same for both horizontal and vertical motion).
Since the ball was thrown at an angle of 30 degrees with the horizontal, you can determine the time it takes for the ball to travel the horizontal distance using the equation:
t = s / (v*cos(theta))
where theta is the angle (30 degrees) and v*cos(theta) is the horizontal component of the velocity.
Substituting the given values, you have:
12.2m = v*t*cos(30)
Now, divide both sides by v*cos(30):
12.2m / (v*cos(30)) = t
Rearrange the equation to solve for v:
v = 12.2m / (t*cos(30))
You need to find the value of t in order to proceed. To do that, you can use the equation for vertical motion. Since the ball has the same height above the ground as the passer's throwing hand, the initial vertical velocity is 0. The equation for vertical motion becomes:
s = (1/2)*a*t^2
where s is the vertical displacement (height), a is the acceleration due to gravity (-9.8m/s^2), and t is the time.
Substituting the given values, you have:
0 = (1/2)*(-9.8m/s^2)*t^2
Simplifying the equation, you get:
t^2 = 0
Since t cannot be zero (as the ball was thrown), this equation gives you no useful information. Therefore, you cannot determine the time using this equation.
So, it looks like there is an issue with part b of this problem. It seems that there must be either missing information or an error in the question itself. Without the time (t), it is not possible to determine the vertical displacement of the ball, and thus we cannot find the answer to part b.