Answer:
v_y = 12.54 m/s
Explanation:
Given values:
- Initial vertical height y_o = 10 m
- Initial velocity v_y,o = 0 m/s
- The object's acceleration in the air = a_y
- The actual time taken to reach the ground t = 3.2 s
Find:
- How to calculate the object's speed when it arrives at the ground?
Solution:
- Apply kinematic equations to find the actual acceleration of the ball when it reaches the ground:
y = y_o + v_y,o*t + 0.5*a_y*t^2
0 = 10 + 0 + 0.5*a_y*(3.2)^2
a_y = - 20 / (3.2)^2 = 1.953125 m/s^2
- Use the total energy conservation principle of the system:
E_p - W_f = E_k
Where, E_p = m*g*y_o
W_f = m*a_y*(y_i - y_f)..... Reflecting air resistance
E_k = 0.5*m*v_y^2
Thus, m*g*y_o - m*a_y*(y_i - y_f) = 0.5*m*v_y^2
g*(10) - (1.953125)*(10) = 0.5*v_y^2
v_y = sqrt(157.1375)
v_y = 12.54 m/s
The approximate answer is 6.84. Reasoning: (2.78^2 + 6.25^2)^(1/2) = 6.84 approx.
Answer:
17 m/s south
Explanation:
Mass of the dog = 10 kg
Mass of skateboard = 2 kg
v = Combined velocity = 2 m/s
Velocity of the dog = 1 m/s
Velocity of skateboard
In this scenario, linear momentum is conserved
The speed of the skateboard post-dog jump will be 17 m/s south, considering north as the positive direction
