The differential equation d^2 y/dt^2 = -g. where g is the gravitational constant

Question

The differential equation d^2 y/dt^2 = -g. where g is the gravitational constant, describes the motion of an object under gravity when air resistance is ignored. Write the general solution for the geographic location of the projectile at time t, i.e. r(t) = (x(t), y(t)) in terms of its initial location (x_0, y_0) and velocity (u_0, v_0). If the object's initial velocity is 50m/s at an angle of 60 degrees up from the horizontal when it is a horizontal distance of 40m from its initial location, it will have a height = and velocity = If the object has the same initial speed as in part(b), and when it is a horizontal distance of 40m from its initial location, its height is 10m. then its initial velocity v_0 = T/F/N: the answer to part (c) would be identical if the differential equation had included a term to account for ah resistance.

Explanation / Answer

(a) Vertical direction
y = Yo + vo t + 0.5gt2
In x direction
x = Xo + uot
(b) Velocity in x direction = 50Cos60 = 25 m/s
  Time = distance / velocity = 40/25 = 1.6 seconds
y = 50Sin60*1.6 + 0.5*(-9.81)1.62  = 56.725 m
Vy = u +at = 50Sin60 -9.81*1.6 = 27.6 m/s
V = (27.62 + 252)1/2 = 37.24 m/s

(c) 40 = 50Cosx*t
10 = 50Sinx*t -0.5gt2
solving 1 and 2 , we get x = 18.645
therfore V0 = 50Sin18.645 = 15.98 m/s
(d) It would be false

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