A projectile is fired at 12.5 m/s at an angle of 53.1 with the horizontal from a point 75m above the ground.

Question

A) How long does it take to reach the ground?
B)What max height does it reach?
C) What horizontal distance does it travel before striking the ground?
D) With what velocity does it strike the ground?

Answer 1

Vo = 12.5m/s[53.1o].
Xo = 12.5*Cos53.1 = 7.51 m/s.
Yo = 12.5*sin53.1 = 10.0 m/s.

A. h = ho + -(Yo^2)/2g=75 + 10^2/19.6 =
80.1 m. Above gnd.

Y = Yo + g*Tr = 0.
Tr = -Yo/g = -10/-9.8 = 1.02 s. = Rise
time.

0.5q*Tf^2 = 80.1 m.
4.9Tf^2 = 80.1.
Tf^2 = 16.34.
Tf = 4.04 s. = Fall time.

Tr+Tf = 1.02 + 4.04 = 5.06 s. = Time to
reach gnd.

B. h = 80.1 m.(Calculated in part A.).

C. Dx = Xo*(Tr+Tf) = 7.51m/s * 5.06s. =
38 m.

D. Y = Yo + g*(Tr+Tf) = 0 + 9.8*5.06 =
49.6 m/s. = Ver. Component of velocity.

V = Sqrt(Xo^2 + Y^2).