Will the dart hit the target? Pre-Calculus velocity help!Maria and Ed are launching yard darts 20 ft from the front edge of a circular target of...

The initial velocity of the dart has the vertical and
horizontal components of 25sin55 and 25 cos 55 feet. The horizontal component remains
constant and the vertical component is affected by the gravitational acceleratin g (=32
feet/sec^2). Consequently. The horizontal and vertical components of displacement at any
time T is  given by the parametric equations

x = uT cos x
...........(1) and

y = uT sinx -(1/2)gT^2 +4 .....(2) where
u is the initial velocity (=25 feet/sec)  of the dart, and T is the time  after release
of the dart. And x is the angle of projection = 55 degree.Clearly when T = 0, y = 4 feet
the height from which Sue releases the dart.

When the dart
hits the ground y = -4 = uTsinx -(1/2)gT^2+4 Or it is a quadratic in T: gT^2-(2usix)T-16
= 0  Or 32T^2 - (2usinx)T -16  = 0. Or 16T^2-(25sin55)T - 8 + 0.
Therefore,

T =  {25sinx55+ sqrt[(25sin55)^2+4*16*8)]}/(32)
=

= 1.593666915  seconds ........(3)when the dart hits
ground.During this time, the Horizontal distance travelled as given by the equation 2
is

x = uTcosx , where the value of T in seconds is as  at
(3)

So x = 25* 1.593666915 cos 55 = 22.85224475
feet.

Therefore, the dart shall fall in the circle of 18
inch as the circle's nearest and farthest cirumference from Sue are at 20 feet and
20feet+ 2radius = 20feet+36" = 20+3 feet =23 feet respectively, provided the vertical
plane of the projectile is passes through the line joining the centre of the cicle and
positon of the dart thrower. Lateral shift of the trough is assumed
zero).

Maria throws the dart afre a time t. So the the dart
assumes the initial postion  of 4 feet at time t. Therefore
T-t

So the equation for the projectile of  Maria's through 
is given bY replacing T in above by T-t.

x = 25(T-t)
cos55.

Your equations are correct ecept a small typo in the
second 2nd eqution.

y
=25(T-t)sin55-(1/2)16(T-t)^2+4