Answer:
a) t =3.349 s
b) V_x,i = 44.8 m/s
c) V_y,f = 32.85 m/s
d) V = 55.55 m/s
Explanation:
Given:
- Total throw in x direction x(f) = 150 m
- Total distance traveled down y(f) = 55 m
Find:
a) How long is the rock in the air in seconds.
b) What must have been the initial horizontal component of the velocity, in meters per second?
c) What is the vertical component of the velocity just before the rock hits the ground, in meters per second?
d) What is the magnitude of the velocity of the rock just before it hits the ground, in meters per second?
Solution:
- Use the second equation of motion in y direction:
y(f) = y(0) + V_y,i*t + 0.5*g*t^2
- V_y,i = 0 (horizontal throw)
55 = 0 + 0 + 0.5*(9.81)*t^2
t = sqrt ( 55 * 2 / 9.81 )
t =3.349 s
- Use the second equation of motion in x direction:
x(f) = x(0) + V_x,i*t
150 = 0 + V_x,i*3.349
V_x,i = 150 / 3.349 = 44.8 m/s
- Use the first equation of motion in y direction:
V_y,f = V_y,i + g*t
V_y,f = 0 + 9.81*3.349
V_y,f = 32.85 m/s
- The magnitude of velocity of ball when it hits the ground is:
V^2 = V_y,f^2 + V_x,i^2
V = sqrt (32.85^2 + 44.8^2)
V = 55.55 m/s
