An airplane with a speed of 85.3 m/s is climbing upward at an angle of 45.0 ° with respect to the horizontal. When the plane's altitude is 587 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.

what I'm doing is that Im finding voy=sin45*85.3=60.32
then i'm finding the time by using the equation vot+0.5at^2
then i'm using the equation D=v*t to find the distance, but i'm getting the wrong answer. pls help clarify the steps and what i'm doing wrong.

Vi = 60.32 yes

u = 85.3 cos 45 = 60.32 also but forever

v = Vi + a t = 60.32 -9.81 t
so at top when v = 0
t = 6.15 seconds coasting up

h at top = Hi + Vi t -4.9 t^2
= 587 + 60.32 (6.15) - 4.9(6.15^2)
= 773 meters high at 6.15 seconds when fall starts

now it falls
773 = 4.9 t^2
t = 12.6 second fall
so
total time after relaese = 12.6+6.15 = 18.7 seconds at horizontal velocity 60.32
d = 18.7 * 60.32 = 1,128 meters

velocity
u = 60.32 we know
v = g t = 9.81*12.6 = 124 down, negative really

speed = sqrt (u^2+v^2) = 138 m/s
angle down from horizontal = A
tan A = v/u = 124/60.32
A = 64 degrees below horizontal

thank you very much Damon. It is much clear now, since u explained it step by step.

To find the distance along the ground, measured from a point directly beneath the point of release, where the package hits the earth, we need to break down the problem into horizontal and vertical components.

(a) Let's first calculate the time it takes for the package to hit the ground. We know that the vertical component of the initial velocity is 60.32 m/s (found correctly) and that the acceleration due to gravity is 9.8 m/s². We can use the equation to calculate the time (t) it takes for the package to hit the ground:

0 = voy - gt

Substituting the given values:

0 = 60.32 - 9.8t

Rearranging the equation:

9.8t = 60.32

t = 60.32 / 9.8 ≈ 6.16 s

Now that we have the time (t), we can find the horizontal distance (D) traveled by the package. The horizontal component of the initial velocity is also 60.32 m/s, and the time is 6.16 s. We can use the equation to calculate the horizontal distance:

D = vox * t

Substituting the given values:

D = (cos 45°) * 85.3 m/s * 6.16 s

Remember to use the cosine function for the horizontal component since it is adjacent to the angle. The cosine of 45° is 0.707.

D ≈ 0.707 * 85.3 * 6.16 ≈ 366.3 m

Therefore, the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth is approximately 366.3 meters.

(b) To find the angle of the velocity vector of the package just before impact, we can use trigonometry. The velocity vector just before impact consists of both horizontal and vertical components. We can find the resultant velocity using the Pythagorean theorem:

v^2 = vx^2 + vy^2

Substituting the known values:

v^2 = (60.32 m/s)^2 + (60.32 m/s)^2

v^2 = 3638.22 m^2/s^2

Taking the square root of both sides:

v ≈ 60.32 m/s

To find the angle (θ) of the velocity vector, we can use the inverse tangent function:

θ = tan^(-1) (vy / vx)

Substituting the known values:

θ = tan^(-1) (60.32 m/s / 60.32 m/s)

θ ≈ tan^(-1) (1) ≈ 45°

Therefore, the angle of the velocity vector of the package just before impact, relative to the ground, is approximately 45°.