A projectile is launced atan angle of 50 degrees with respect to the horizontal

Question

A projectile is launced atan angle of 50 degrees with respect to the
horizontal at an initial speed of 28.0 m/s.
The projectile impacts into a tall building some distance from thelaunch.
The bases of the building is at the same height as the launchpoint. When the
projectile impacts it velocity vector has an angle of 3.degrees
below the horizontal.

(a) Find the initialhorizontal and vertical components of the
projectile's velocity.

(b) Find the horizontaland vertical components of the projectile's
velocity 1.2 seconds before it reaches the heighest of itstrajectory. ( point A)

(c) what is the velocityand acceleration of the projectile when it is at ppointA?

(d) What is the verticalcomponent of the projectile's velocity just before impact? ( Inm/s)

Explanation / Answer

Part A) We Know: Initial Velocity (Vo) = 28 m/s Angle of Trajectory () = 50° Solve for x/y velocities: Velocity in the y-direction (Vy) =Vosin   = 28sin50 = 21.45m/s Velocity in the x-direction (Vx) =Vocos = 28cos50 = 18m/s Part B) Horizontal Component (x-direction): Velocity remains constant in the x-direction (18m/s) assuming there is no air resistance Vertical Component (y-direction): V(t) = Vy + at    ----->where    a = -9.8m/s² (gravity)    t= 1.2s (time) V(t) = velocity at time t (Unknown) V(1.2s) = 21.45 m/s + (-9.8m/s²)(1.2s) V(1.2s) = 9.69 m/s in the y-direction after 1.2seconds Part C) Velocity in the y-direction at the peak of the trajectory (point A)= 0 m/s             (it is stopped in theair for an instant and is about to fall down) Velocity in the x-direction at the peak of the trajectory (point A)= 18m/s         (velocity in thex-direction is constant as before assuming no air resistance) Acceleration acts only in the y-direction and is constant at-9.8m/s² throughout the trajectory Part D) To find the height just before impact Use the followingformula: tan(30°) = Vy(at 30°) / (Vx) tan(30°) = Vy(at 30°) / (18m/s) Vy(at 30°) = (18m/s)tan(30°) Vy(at 30°) = 10.39 m/sdownward (so it's direction is negative) Part E) Find time it takes to reach peak: Vp = Vy + at 0 m/s = 21.45m/s + (-9.8)(t) t = 2.19 s Find time it takes from peak to point of impact: Vi = Vp + at -10.39m/s = 0 + (-9.8m/s)(t) t = 1.06 s Add the times together: 2.19 s + 1.06 s = 3.25 seconds tillobject reaches point of impact

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