Respuesta :
y = 0m
y0 = 166m
v0y = 0 m/s
g = 9.8 m/s^2
t = ?
Solve for t:
y = y0 + v0y*t - (0.5)gt^2
0 = 166 - (0.5)(9.8)t^2
t = 5.82 s
Now, using time, we can solve for the range using the equation:
x = vx(t)
x = (40)(5.82)
x = 232.8 m
The impact horizontal component of velocity will be 40 m/s as velocity in terms of x is always constant. To find the impact vertical component of velocity, we use the equation:
v = v0y - gt
v = 0 - (9.8)(5.82)
v = -57.04 m/s
Answer:
The package will strike at 247.65 m.
Explanation:
Given that,
Speed = 39 m/s
Height = 198 m
Acceleration = 9.8 m/s²
We need to calculate the time
Using equation of motion
[tex]s = ut+\dfrac{1}{2}gt^2[/tex]
Put the value into the formula
[tex]198=0+\dfrac{1}{2}\times9.8\times t^2[/tex]
[tex]t=\dfrac{198\times2}{9.8}[/tex]
[tex]t=\sqrt{40.40}[/tex]
[tex]t=6.35\ sec[/tex]
We need to calculate the distance where the package strike the ground
Using formula of distance
[tex]d= vt[/tex]
Put the value into the formula
[tex]d=39\times6.35[/tex]
[tex]d=247.65\ m[/tex]
Hence, The package will strike at 247.65 m.
