A rock is thrown off a cliff with a vertical speed of 53 m/s upward and a horizontal speed of 30 m/s.
A) what will these speeds Vx and Vy, be 4 seconds later?
B) what will be the x and y displacements of the rock be 4 seconds later?
Vx is constant at 30
Vy = 53-9.8t
x(t) = 30*t
y(t) = 53t - 4.9t^2
To answer these questions, we need to consider the motion of the rock in both the horizontal and vertical directions separately.
A) To find the vertical and horizontal speeds 4 seconds later, we can use the following equations:
Vertical speed (Vy) at time t = initial vertical speed (Vy0) - acceleration due to gravity (g) x time (t)
Horizontal speed (Vx) at time t = initial horizontal speed (Vx0)
Given:
Vy0 = 53 m/s (upward) ⇒ Since we are taking upward direction as positive, Vy0 = +53 m/s
Vx0 = 30 m/s
g = 9.8 m/s^2 (acceleration due to gravity)
t = 4 s
Vertical speed (Vy) at time t = 53 m/s - (9.8 m/s^2 x 4 s) = 53 m/s - 39.2 m/s = 13.8 m/s (upward)
Horizontal speed (Vx) remains the same: Vx = 30 m/s
Therefore, after 4 seconds:
Vertical speed (Vy) = 13.8 m/s (upward)
Horizontal speed (Vx) = 30 m/s
B) To find the x and y displacements (horizontal and vertical positions) after 4 seconds, we can use the following equations:
Vertical displacement (Δy) = initial vertical speed (Vy0) x time (t) + (1/2) x acceleration due to gravity (g) x time (t)^2
Horizontal displacement (Δx) = initial horizontal speed (Vx0) x time (t)
Given:
Vy0 = +53 m/s
Vx0 = 30 m/s
g = 9.8 m/s^2 (acceleration due to gravity)
t = 4 s
Vertical displacement (Δy) = (+53 m/s) x (4 s) + (1/2) x (9.8 m/s^2) x (4 s)^2
= 212 m + 78.4 m
= 290.4 m (upward)
Horizontal displacement (Δx) = (30 m/s) x (4 s)
= 120 m
Therefore, after 4 seconds:
Vertical displacement (Δy) = 290.4 m (upward)
Horizontal displacement (Δx) = 120 m