Question

A motorcycle stunt rider rides off the edge of a cliff. Just at the edge his velocity is horizontal, with the magnitude of 9.0 m/s. After 0.50s, find...

[Answer questions a, b, c, d, & e below. Show all work and details. Give explanations to support your answers]

a) the motorcycle's position

b) distance from the edge of the cliff.

c) velocity.

d) speed

e) direction with repect to the horizontal direction

Answer 1

(a) For X - direction -  
Vi = 9.0 m/s
Vf = ?
x = ?
a = 0 m/s²
t = 0.50 s

y-direction:
Vi = 0 m/s
Vf = ?
y = ?
a = 9.8 m/s²
t = 0.50 s

equations we will use:
x = Vi * t + (1/2) * a * t²
Vf = Vi + a* t

we can find the horizontal position by using the first equation with the x-direction information:
x = (9)(0.50) + (1/2)(0)(0.50)²
x = (9)(0.50) + 0
x = 4.50 m

to find the vertical position, use the first equation with the y-direction information:
y = (0)(0.50) + (1/2)(9.8)(0.50)²
y = 0 + (1/2)(9.8)(0.25)
y = 1.225 m

actual position = ?(x² + y²)
?[(4.5)² + (1.225)²] = 4.66 m  

tan(?) = y / x
? = tan?¹(y / x)
? = tan?¹(1.225 / 4.50) = 15.23° south of east

So, motorcycle's position is 4.66 m, 15.23° south of east

(b) Distance from the edge of the cliff = 1.225 m

(c) To find the riders velocity we will use the second equation with the y-information:

Vf = Vi + a * t
Vf = (0) + (9.8)(0.50) = 4.9 m/s

and now with the x information:
Vf = Vi + a * t
Vf = (9) + (0)(0.25) = 9.0 m/s

actual velocity = ?(x² + y²)
?(4.9² + 9²) = 10.25 m/s

(d) Speed is also 10.25 m/s

(e) theta = tan^-1(Vy / Vx) = tan^-1(4.9 / 9) = 28.6 deg.