Question

An airplane with a speed of 90.8 m/s is climbing upward at an angle of 54.1 ° with respect to the horizontal. When the plane's altitude is 836 m, the pilot releases a package. (a)Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth. (b) Relative to the ground, determine the angle of the velocity vector of the package just before impact.

Answer 1

v = 90.8 m/s
θ = 54.1 ^o
h = 836 m
vx = 90.8 * cos(54.1) = 53.24 m/s
vy = 90.8 * sin(54.1) = 73.55 m/s

time taken to reach ground be t
- 836 = 73.5 * t - 1/2*9.8*t^2
t = 22.56 s

Distance travelled in horiontal direction, sx = vx * t
sx = 53.24 * 22.56
sx = 1201 m

(b)
Calculating the vertical component of velocity just before it hits the ground,vyf

vyf = v + a*t
vyf = 73.55 - 9.8 * 22.56
vyf = - 147.5 m/s

vx = 53.24 m/s

tan(θ) = vyf/vx
θ = tan^-1(  - 147.5/53.24)
θ = - 70.15^o

-ve sign shows , angle is below the horizontal !!