A bomber in level flight, flying at 92m/s, releases a bomb at a height of 1950m.
(a) How long is it before the bomb strikes earth? (19.9s)
(b) How far does it travel horizontally? (1832m)
(c) What are the horizontal and vertical components of its velocity when it strikes? (195m/s)
(d) What is its velocity of impact? (216m/s)
i understand a and b , i just need help with c and d.
can u show all ur steps please, thank you!
horizontal is constant at 92 m/s
vertical is 9.8*19.9
so, (d) is just the resultant speed s, where s^2 = h^2 + v^2
Sure, I'd be happy to help you with parts (c) and (d). Let's break it down step by step:
(c) To find the horizontal and vertical components of velocity when the bomb strikes, we need to determine the components of its initial velocity. The initial velocity can be split into two parts: the horizontal component and the vertical component.
Given:
Initial velocity in level flight = 92 m/s
Angle of descent (since it's in level flight) = 0 degrees (horizontal)
Angle of descent (since it's descending) = 90 degrees (vertical)
To find the horizontal component of velocity:
Horizontal velocity (Vx) = Initial velocity * Cos(angle of descent)
Vx = 92 m/s * Cos(0 degrees)
Vx = 92 m/s * 1 (Cos(0 degrees) = 1)
Vx = 92 m/s
So, the horizontal component of the velocity when the bomb strikes is 92 m/s.
To find the vertical component of velocity:
Vertical velocity (Vy) = Initial velocity * Sin(angle of descent)
Vy = 92 m/s * Sin(90 degrees)
Vy = 92 m/s * 1 (Sin(90 degrees) = 1)
Vy = 92 m/s
So, the vertical component of the velocity when the bomb strikes is 92 m/s.
(d) To find the velocity of impact, we can use the Pythagorean theorem to combine the horizontal and vertical components of velocity:
Velocity of impact (V) = √(Vx^2 + Vy^2)
V = √(92 m/s)^2 + (92 m/s)^2
V = √(8464 m^2/s^2 + 8464 m^2/s^2)
V = √(16928 m^2/s^2)
V = √16928 m^2/s^2
V ≈ 130 m/s
So, the velocity of impact when the bomb strikes is approximately 130 m/s.
I hope this helps! Let me know if you have any further questions.
Certainly! I'd be happy to help you with parts (c) and (d). Here are the step-by-step solutions:
(c) To find the horizontal and vertical components of the bomb's velocity when it strikes, we can use the equations of motion.
We know that the initial vertical velocity is zero, as the bomb is released in level flight. The acceleration due to gravity is -9.8 m/s², considering downwards as negative.
Using the equation of motion for displacement in the vertical direction:
Δy = v_iy × t + (1/2) × a_y × t²
where Δy is the vertical displacement, v_iy is the initial vertical velocity, t is the time, and a_y is the vertical acceleration.
We can rewrite this equation as:
-1950m = 0 × t + (1/2) × (-9.8 m/s²) × t²
-1950m = -4.9 m/s² × t²
Now, let's solve for t:
4.9 m/s² × t² = 1950 m
t² = 1950 m / (4.9 m/s²)
t² = 400
t = √(400)
t = 20 s (taking the positive value as time can't be negative in this case)
Now, let's find the horizontal displacement using the equation:
Δx = v_ix × t
where Δx is the horizontal displacement and v_ix is the initial horizontal velocity.
Given that the bomber is in level flight, the horizontal velocity remains constant throughout. So, v_ix = 92 m/s.
Δx = 92 m/s × 20 s
Δx = 1840 m
Therefore, the bomb travels approximately 1840 m horizontally.
(d) To find the velocity of impact, we can use the Pythagorean theorem to find the magnitude of the final velocity.
The final velocity (v_f) is given by:
v_f = √(v_xf² + v_yf²)
where v_xf is the horizontal component of the velocity at impact and v_yf is the vertical component of the velocity at impact.
We know that v_yf = -9.8 m/s (since it is the acceleration due to gravity) and v_xf = 92 m/s (as the horizontal velocity remains constant).
v_f = √(92 m/s)² + (-9.8 m/s)²
v_f = √(8464 m²/s² + 96.04 m²/s²)
v_f = √(8560.04 m²/s²)
v_f = 92.59 m/s (rounded to two decimal places)
Therefore, the velocity of impact is approximately 92.59 m/s.
I hope this helps! Let me know if you have any further questions.
Sure! Let's go through steps for solving parts (c) and (d).
(c) To find the horizontal and vertical components of the velocity when the bomb strikes, we can use the concept of projectile motion. When the bomb is released from the bomber, it will have both a horizontal and a vertical velocity component.
The horizontal velocity component remains the same throughout the flight since there is no force acting horizontally on the bomb. Therefore, the horizontal component of the velocity remains constant at 92 m/s.
The vertical velocity component changes due to the effect of gravity. The initial vertical velocity of the bomb is 0 m/s when it is released. We can use the equation for vertical motion in free fall to find the final vertical velocity when the bomb strikes:
v_f = v_i + gt
where:
v_f is the final vertical velocity (unknown)
v_i is the initial vertical velocity (0 m/s)
g is the acceleration due to gravity (-9.8 m/s^2) (taking downward as negative)
t is the time taken for the bomb to reach the ground (which we found in part (a), 19.9 s)
Plugging in the values, we get:
v_f = 0 + (-9.8) * 19.9 = -195 m/s
The negative sign indicates that the final vertical velocity is directed downwards, which is consistent with the motion of the bomb.
So, the horizontal component of the velocity when the bomb strikes is 92 m/s (same as before), and the vertical component is -195 m/s.
(d) The velocity of impact is the resultant velocity of the horizontal and vertical components. To find this, we can use the Pythagorean theorem:
v_impact = √(v_horizontal^2 + v_vertical^2)
Plugging in the values, we get:
v_impact = √(92^2 + (-195)^2) = √(8464 + 38025) = √46489 = 215.7 m/s (approx. 216 m/s)
Therefore, the velocity of impact is approximately 216 m/s.
I hope this helps! Let me know if you have any further questions.