A plane, diving with constant speed at an angle of 51.5° with the vertical, releases a projectile at an altitude of 654 m. The projectile hits the ground 6.40 s after release. (a) What is the speed of the plane? (b) How far does the projectile travel horizontally during its flight? What were the magnitudes of the (c) horizontal and (d) vertical components of its velocity just before striking the ground? (State your answers to (c) and (d) as positive numbers.)
h= v(oy) +gt²/2
v(oy)=( h/t )– (gt/2) = 7.83 m/s.
v(oy) =v(o) •cosα .
(a)
v(o) = v(oy)/cosα = 70.83/cos51.5 = 113.8 m/s.
(b) v(ox)=v(x) = v(o)sinα=
=113.8•sin51.5 =89.06 m/s
s=v(x) •t =89.06•6.4=570 m.
(c)
v(y) =v(oy) +gt =7.83 +9.8•6.4=70.55 m/s.
v(x) =89.06 m/s
To solve this problem, we can break it down into several steps:
Step 1: Determine the initial vertical velocity of the projectile
Since the plane is diving at an angle of 51.5° with the vertical, the vertical component of its velocity will be Vsin(51.5°). We need to find the value of V, the speed of the plane, in order to determine the initial vertical velocity of the projectile.
Step 2: Find the time it takes for the projectile to reach the ground
We are given that the projectile hits the ground 6.40 seconds after release. So the time it takes for the projectile to reach the ground is 6.40 s.
Step 3: Calculate the initial vertical velocity of the projectile
To find the initial vertical velocity of the projectile, we need to consider its motion starting from the moment it is released. The time it takes for the projectile to reach the ground is the same as the time it takes for the plane to travel a horizontal distance. So, we can use this time to find the initial vertical velocity.
The vertical distance traveled by the projectile is given as 654 m. So, we can use the equation:
(1/2)at^2 + V_y*t + y_0 = 0,
where a is the acceleration due to gravity (-9.8 m/s^2), t is the time (6.40 s), V_y is the initial vertical velocity, and y_0 is the initial vertical position (654 m).
Rearranging the equation, we get:
(1/2)(-9.8)(6.40)^2 + V_y * (6.40) + 654 = 0
Now, we can solve this equation to find the value of V_y, which is the initial vertical velocity of the projectile.
Step 4: Calculate the initial horizontal velocity of the projectile
The horizontal distance traveled by the projectile can be calculated using the equation:
Horizontal distance = (initial horizontal velocity) * (time)
The initial horizontal velocity is the same as the velocity of the plane, V. We can now calculate the horizontal distance traveled by the projectile.
Step 5: Determine the velocity components just before striking the ground
To find the magnitudes of the horizontal and vertical components of the velocity of the projectile just before striking the ground, we need to consider the horizontal and vertical velocities separately.
The horizontal component remains constant throughout the flight. Hence, the horizontal component just before striking the ground is the same as the initial horizontal velocity.
The vertical component changes due to the acceleration due to gravity. We can calculate the vertical component just before striking the ground by using the equation:
Final vertical velocity = initial vertical velocity + (acceleration due to gravity * time)
In this case, the final vertical velocity just before striking the ground will be -9.8 m/s (since the projectile is moving downward).
Now, we have all the information required to solve the problem.
(a) To find the speed of the plane (V):
V = V_x / cos(θ),
where V_x is the horizontal component of the plane's velocity and θ is the angle of the plane with the vertical.
(b) The horizontal distance traveled by the projectile is given as the product of the initial horizontal velocity and the time taken.
(c) The horizontal component of the velocity just before striking the ground is equal to the initial horizontal velocity.
(d) The vertical component of the velocity just before striking the ground is the final vertical velocity, which is -9.8 m/s.