PHYSICS HELP PLS?

Question

In a certain biathlon competition the athlete must hit a target using a blowgun after taking a physics exam for 3 hours. A blowgun is a tube in which a dart is sent out by blowing on one end. The target is a 1.0 cm diameter circle at 5.00 m height on a wall 10.0 m from the end of the tube, as in the diagram below. The inexperienced athlete lines up the tube of the blowgun pointing directly at the target, such as along the dotted line shown in the 
diagram. They blow a dart with an initial speed of 35.0 m/s. Assuming no air resistance: 
a) Where does the dart hit the wall? Show a plot of the projectile y(x) graph indicating the important parameters. (5 pts) 
b) Given the same speed as in (a), what must the initial angle of the dart be in order to hit the centre of the target? (4 pts) 
c) If there was air resistance, what effects would you expect to have on your answers 
above? (1) 
(Note: A rifle which shoots bullets at much higher speeds will have calibrated the angle of the aiming sites along the rifle barrel to partially compensate for the effects of gravity and air.)

The prof has said knowing this trig identity may be helpful : 1/cos^2x= 1+tan^2x

But I don't even know how to approach the problem! An explanation would be helpful.

? · Accepted Answer

s_vert = y = s0 + ut + (a/2)t^2
y = 5 - 4.905t^2

s0 = 5 height of target and blowgun
u = initial velocity (vertically = 0)
a = -9.81 m/s^2 (vertical acceleration downward due to gravity)

horiz
s = 10 = 0 + 35t
t = 0.286 s

vert
s = 5 - 4.905(0.286)^2
s = 4.6 m (height on wall)

horiz
10 = 35 cos x t
t = 0.286 / cos x

s = 5 = 5 + 35 sin x (0.286 / cos x) - 4.905(0.286 / cos x)^2
0 = 10.01 (sin x / cos x) - 0.401 / cos^2 x
0 = 10.01 tan x - 0.401 - 0.401 tan^2 x
quadratic formula in terms of tan x 
solving yields tan x = 0.040124 or 24.92247
x = 2.3 or 87.7 degrees above horizontal
x = 2.3 degrees (checking 87.7 reveals it is extraneous root)

dart would hit wall lower in (a) due to horizontal -ve acceleration, thus increased time, thus increased gravitation acceleration downward
angle would need to increase in (b) to compensate for increased drop

PHYSICS HELP PLS?

1 Answer