A plane flying horizontally at 180 m/s drops a box just as it crosses a coastline. The box lands 3 kilometers from the coast.
- At what height is the plane flying?
- How long it takes to the box to reach the ground ?
time in air: 3000/180 seconds
heightItFell=1/2 g t^2
5 m
To find the height at which the plane is flying, we can use the equation for projectile motion:
y = y0 + v0y*t - (1/2)*g*t^2
where:
y = height
y0 = initial height
v0y = vertical component of the initial velocity
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Since the plane is flying horizontally, the initial height (y0) can be assumed to be zero. The vertical component of the initial velocity (v0y) can be found using the equation:
v0y = v0 * sin(theta)
where:
v0 = initial velocity of the plane (180 m/s)
theta = angle of elevation of the plane (which is assumed to be zero since the plane is flying horizontally)
Substituting the values into the equation, we have:
y = 0 + (180 m/s * sin(0))*t - (1/2)*9.8 m/s^2 * t^2
Simplifying the equation, we have:
y = 0 + 0*t - 4.9 t^2
Since the height (y) is zero when the box touches the ground, we can solve for t:
0 = - 4.9 t^2
Using the quadratic formula, we find that t = 0 and t = sqrt(0/(-4.9)), which simplifies to t = 0 and t = 0.
Since the time cannot be zero, the box takes no time to reach the ground. Therefore, it is already on the ground when it lands 3 kilometers from the coast.
To find the height at which the plane is flying, we substitute this time value (t = 0) into the equation for height:
y = 0 + 0*0 - 4.9*(0)^2
Therefore, the height at which the plane is flying is 0 meters.