Answer:
133.9 ft/s
Explanation:
We can find the horizontal component and the vertical component of the initial velocity by considering the horizontal and vertical motions indipendently.
Along the horizontal direction, the rock travels at constant speed: this means that its horizontal velocity is constant, and it is given by
[tex]u_x = \frac{d}{t}[/tex]
where
d = 160 ft is the distance covered
t = 5.0 s is the time taken
Substituting,
[tex]u_x = \frac{160}{5}=32 ft/s[/tex]
Along the vertical direction, the rock is in free-fall - so its motion is a uniform accelerated motion with constant acceleration g = -32 ft/s^2 (downward). Therefore, the vertical distance covered is given by the suvat equation:
[tex]s=u_y t + \frac{1}{2}gt^2[/tex]
where
s = -100 ft is the vertical displacement
[tex]u_y[/tex] is the initial vertical velocity
Replacing t = 5.0 s and solving the equation for [tex]u_y[/tex], we find:
[tex]u_y = \frac{s}{t}-\frac{1}{2}gt=\frac{100}{5}-\frac{1}{2}(-32)(5)=130 ft/s[/tex]
Therefore, the initial speed is just the magnitude of the initial velocity:
[tex]u=\sqrt{u_x^2+u_y^2}=\sqrt{32^2+130^2}=133.9 ft/s[/tex]
Answer:
68ft
Explanation:
x-direction
Δx = vt
160 = v(5)
v = 32
y-direction
Δx = [tex]v_{0y} t + \frac{1}{2} a_{y} t^{2}[/tex]
*note that when using ft, acceleration by gravity is -32, not -9.8
-100 = v(5) + 0.5(-32)(5^2)
v = 60
adding horizontal & vertical vectors to find resultant
use pythagreon theorem
32^2 + 60^2 = h^2
h = 68
