A bomber, flying upwards at an angle 53 with verticql releases a bomb at 800m altitude. The bomb strikes the ground 20s after release. Find
A--velocity of bomber at time of release of bomb
B--maximum height attained by bomb
C--horizontal distance covered by bomb before reaching gorund
D--velocity of bomb when it strikes the ground
To solve this problem, we'll use the equations of motion for projectiles. Let's start with part A.
A) Velocity of the bomber at the time of release of the bomb:
The velocity of the bomber at the time of release can be split into two components: horizontal and vertical.
The horizontal component remains constant throughout the motion, while the vertical component decreases due to gravity.
The initial vertical velocity can be calculated using the formula:
V_y = V * sin(θ)
where V is the initial velocity of the bomber and θ is the angle at which the bomber is flying.
Given that θ = 53 degrees, we need to find V.
Next, let's calculate the time it takes for the bomb to hit the ground.
B) Time of flight:
The time of flight is the time it takes for the bomb to reach the ground. We can find it using the formula:
t = 2 * (V_y) / g
where g is the acceleration due to gravity.
Given that the time of flight is 20 seconds, we can solve for V_y.
Now that we have both V_y and V, we can find the initial velocity of the bomber using the formula:
V = sqrt((V_x)^2 + (V_y)^2)
where V_x is the horizontal component of the velocity.
Next, let's move on to part B.
B) Maximum height attained by the bomb:
To find the maximum height attained by the bomb, we need to calculate the vertical distance traveled during the time of flight.
Using the formula:
h = V_y * t - 0.5 * g * t^2
where h is the maximum height attained, V_y is the initial vertical velocity of the bomb, and t is the time of flight.
Now, let's calculate the horizontal distance covered by the bomb before reaching the ground in part C.
C) Horizontal distance covered by the bomb:
The horizontal distance can be found using the formula:
d = V_x * t
where d is the horizontal distance, V_x is the horizontal component of the velocity, and t is the time of flight.
Finally, let's determine the velocity of the bomb when it strikes the ground in part D.
D) Velocity of the bomb when it strikes the ground:
The final velocity of the bomb just before it hits the ground can be found using the formula:
V_f = sqrt((V_x)^2 + (V_y)^2 + 2 * g * h)
where V_f is the final velocity of the bomb, V_x is the horizontal component of the velocity, V_y is the initial vertical velocity, g is the acceleration due to gravity, and h is the maximum height attained by the bomb.
Now, let's solve the problem step-by-step to find the answers.
To solve this problem, we can apply basic kinematic equations for projectile motion. Here's how to find each parameter:
A) Velocity of the bomber at the time of release of the bomb:
The vertical velocity component of the bomber doesn't change since the motion is perpendicular to gravity, so it will be the same throughout.
The initial vertical velocity component (Vy_initial) can be found using trigonometry:
Vy_initial = velocity * sin(angle)
Since we are given the angle of 53 degrees, we can use this formula to find Vy_initial.
B) Maximum height attained by the bomb:
To find the maximum height, we need to calculate the time it takes for the bomb to reach its peak. After reaching the peak, the vertical velocity component becomes zero.
We can use the equation of motion to find the time taken for the bomb to reach its peak:
Vy_final = Vy_initial + (g * t)
where g is the acceleration due to gravity (approximately -9.8 m/s²) and t is the time taken to reach the peak.
Solving this equation, we can find the time taken to reach the peak. Then, we can substitute this time into the equation to find the maximum height (H_max):
H_max = Vy_initial * t + (1/2) * (g) * (t)^2.
C) Horizontal distance covered by the bomb before reaching the ground:
The horizontal distance traveled by the bomb (D) can be calculated using the formula:
D = Vx * t
We need to find the horizontal component of the velocity (Vx) of the bomber at the time of release. Since the horizontal velocity is constant, we can use basic trigonometry to find Vx:
Vx = velocity * cos(angle)
D) Velocity of the bomb when it strikes the ground:
To find the vertical velocity component of the bomb (Vy_final) when it strikes the ground, we can use the equation of motion:
Vy_final = Vy_initial + (g * t)
Since we have already calculated the time taken to reach the maximum height, we can substitute this time into the equation to find Vy_final.
Please provide the value of the velocity (in m/s), and we can calculate these parameters accordingly.
Madharchod h Tu sala bhosdi
vertical:
hf=hi+vi*sin53*20-4.9*20^2
hf=0 hi=800
solve for vi
Horizontal:
horiz distance=vi*cos53*20
max height:
mgh=1/2 m (vi*sin53)^2
solve for h, then add initial 800m altitude.
Velocity at impact
1/2 m vi^2+mg*800=1/2 m vf^2
solve for vf
your teacher is too easy.