2D Motion....launch speed...?

Question

I'm really confused about this problem. I have broken it down into components (x, x initial, etc.), but I seem to not have enough info. If someone could please assist me, I would greatly appreciate it. Thanks!

A coyote is chasing a roadrunner. While running down the road, they come to a deep gorge, 15 m straight across and 110 m deep. The roadrunner launches itself across the gorge at a launch angle of 11 degrees above the horizontal, and lands with 2.5 m to spare. The acceleration of gravity is 9.81 m/s^2.

What was the roadrunner's launch speed? Ignore air resistance. Answer in units of m/s.

Answer 1

The two equations that matter are
the horizontal displacement
x(t)=cos(th)*v*t
the vertical displacement
y(t)=sin(th)*v-.5*g*t^2

from the post
112.5=cos(th)*v*t
and
0=sin(th)*v*t-.5*g*t^2

th =11

you can now solve for t and v


solving for v*t first
112.5=cos(11)*v*t

v*t=114.6

plug this into the vertical displacement
0=sin(11)*114.6-.5*9.81*t^2
solve for t

t=2.11 seconds

so v=114.6/2.11

v=54.31 m/s

j

Answer 2

Assuming that it travels in the time of horizontal floor, which you have stated is real, the ball will land with a similar velocity because it took off at, on a similar perspective from the horizontal. you mustn't only be plugging into equations; think of approximately it logically. there isn't any horizontal acceleration, so Vix=Vfx. The vertical acceleration is continuous, and it spends a million/2 of its time going up, being decelerated; to that end, it has a similar volume of time to strengthen up downward after the top of its ascent. it would desire to have a similar velocity. notice that this basically works if the appropriate vertical place is a similar using fact the preliminary vertical place.

Answer 3

Break into vertical and horizontal components.

d^2x/dt^2 = 0
d^2y/dt^2 = -g

dx/dt = V_x = V_i*cosd(11) (cosd is the cosine in degrees so I dont forget)

dy/dt = -g*t + V_y = -g*t + V_i*sind(11)

x(t) = V_i*cosd(11)*t
y(t) = -g/2*t^2 + V_i*sind(11)*t

we know that at the final time, x(tf) = 15+2.5m = 17.5m
x(tf) = V_i*.9816*tf = 17.5m
we also know that y(t) must be zero if he did not fall into the gorge
y(tf) = -g/2*tf^2 + V_i*.1908*tf = 0

solving y(tf) first,
tf = 0 or V_i*.1908*2/g
since tf>0, then tf=V_i*.1908*2/g
substitute this into x(tf)

x(tf) = V_i*.9816*V_i*.1908*2/g = 17.5m

or V_i^2 = 458.3151 (m/s)^2

so V_i = 21.4083 m/s

the guy above's answer is only valid if the gorge is 110 m across! but its only 15, so mine is right.