# 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 radius 18 in. on the ground. Sue releases the dart 4...

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 radius 18 in. on the ground. Sue releases the dart 4 ft above the ground with initial velocity of 25ft/sec at a 55˚ angle.

1. Write parametric equations for the horizontal and vertical distances that the dart is from Maria after

t seconds.

My attempt:

y=-16t^2 + [25sin(55)]t+4

x=[55cos(55)]t

i however dont know how to do part b? help!

neela | High School Teacher | (Level 3) Valedictorian

Posted on

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