Answer:
y₀ = 1020.3 m
Explanation:
This is a projectile launching exercise, in this case as the package is released its initial vertical velocity is zero.
y = y₀ + t - ½ g t²
when it reaches the ground its height is zero
0 = y₀ + 0 - ½ g t²
y₀ = ½ g t²
let's calculate
y₀ = ½ 9.8 14.43²
y₀ = 1020.3 m
Answer:
When have passed 3.9[s], since James threw the ball.
Explanation:
First, we analyze the ball thrown by James and we will find the final height and velocity by the time two seconds have passed.
We'll use the kinematics equations to find these two unknowns.
Note: The sign for the gravity is minus because it is acting against the movement.
Now we can find the velocity after 2 seconds.
Note: The sign for the gravity is minus because it is acting against the movement.
Now we can take these values calculated as initial values, taking into account that two seconds have already passed. In this way, we can find the time, through the equations of kinematics.
As we can see the equation is based on Time (t).
Now we can establish with the conditions of the ball launched by David a new equation for y (elevation) in function of t, then we match these equations and find time t
Then the time when both balls are going to be the same height will be when 2 [s] plus 1.9 [s] have passed after David throws the ball.
Time = 2 + 1.9 = 3.9[s]