A plane, diving with constant speed at an angle of 52.0° with the vertical, releases a projectile at an altitude of 790 m. The projectile hits the ground 6.00 s after release. (Assume a coordinate system in which the airplane is moving in the positive horizontal direction, and in the negative vertical direction. Neglect air resistance.)
(a) What is the speed of the aircraft?
(b) How far did the projectile travel horizontally during its flight?
(c) What were the horizontal and vertical components of its velocity just before striking the ground?
An initial velocity of the projectile v₀ = the velocity of the plane.
v₀ (x) = v₀cosα,
v₀ (y) = v₀sinα,
h= v₀ (y) •t+ gt²/2=
=v₀sinα•t + gt²/2.
v₀={h- gt²/2}/ sinα•t=
={790 – (9.8•36/2)}/sin52•6 =129 m/s
s= v(x)t= v₀cosα•t=790•cos52•6=479.4 m.
v(x) = v₀ (x) = v₀cosα= 129 •cos52 =79.4 m/s
v(y) = v₀ (y)+gt = v₀sinα +gt =
=160.5 m/s
To calculate the speed of the aircraft, we need to find the horizontal component of its velocity and divide it by the cosine of the angle of 52.0°.
To find the horizontal component of velocity, we can use the equation v_x = d_x / t, where v_x is the horizontal velocity, d_x is the horizontal distance traveled, and t is the time taken.
Since the plane is diving at a constant speed, the horizontal velocity does not change. Therefore, we can use the equation v_x = d_x / t, where d_x is the horizontal distance traveled by the projectile.
Given that the projectile hits the ground 6.00 s after release, we need to find the horizontal distance traveled by the projectile during this time.
The horizontal distance traveled by the projectile can be found using the equation d_x = v_x * t.
(a) To find the speed of the aircraft, we can use the equation v_aircraft = v_x / cos(52.0°).
(b) To find the horizontal distance traveled by the projectile, substitute the values into the equation d_x = v_x * t.
(c) To find the horizontal and vertical components of velocity just before striking the ground, we use the equation v_x_final = v_x and v_y_final = v_y_initial + a_y * t, where v_x_final is the final horizontal velocity, v_y_initial is the initial vertical velocity (0 m/s), a_y is the acceleration due to gravity (-9.8 m/s^2), and t is the time taken for the projectile to hit the ground.