Answer:
The magnitude of the stone velocity before it hits the ground is 30 m/s.
Explanation:
Given;
initial vertical velocity, [tex]V_y_i = 10 \ m/s[/tex]
final vertical velocity, [tex]V_y_f[/tex] = 30 m/s
Apply the following kinematic equation to determine the height of the cliff;
[tex]V_y_f^2 = V_y_i^2 + 2gh\\\\30^2= 10^2 + 2(9.8)h\\\\900 = 100 + 19.6h\\\\19.6h = 900 - 100\\\\19.6h = 800\\\\h = \frac{800}{19.6}\\\\h = 40.82 \ m[/tex]
Determine time of the journey;
h = vₓ + ¹/₂gt²
vâ‚“ = 0 (initial horizontal velocity when the stone was thrown vertically downward)
h = ¹/₂gt²
t = √ (2h / g)
t = √ (2(40.82) / 9.8)
t = 2.89 s
When the rock is projected horizontally, the horizontal distance, x = 40.82 m
Initial horizontal velocity, [tex]V_x_i = 10 \ m/s[/tex]
The final vertical component of the velocity is given by;
[tex]V_f_y = V_y_i + gt\\\\V_f_y = 0 + 9.8(2.89)\\\\V_f_y = 28.32 \ m/s[/tex]
The magnitude of the stone velocity before it hits the ground is given by;
[tex]V_f = \sqrt{V_x_i^2 + V_y_f^2} \\\\V_f = \sqrt{10^2 + 28.32^2}\\\\V_f = 30 \ m/s[/tex]
Therefore, the magnitude of the stone velocity before it hits the ground is 30 m/s.
