Home
Class 12
PHYSICS
An airplane, diving at an angle of 53.0^...

An airplane, diving at an angle of `53.0^@` with the vertical releases a projectile at an altitude of 730 m. The projectile hits the ground 5.00 s after being released. What is the speed of the aircraft?

A

`282ms^(-1)`

B

`282ms(-1)`

C

`182ms(-1)`

D

`102ms(-1)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the speed of the aircraft, we can break down the problem step by step. ### Step 1: Understand the problem We have an airplane diving at an angle of 53 degrees with the vertical. It releases a projectile from an altitude of 730 m, and the projectile hits the ground after 5 seconds. We need to find the speed of the aircraft. ### Step 2: Identify the components of motion The motion of the projectile can be analyzed in two dimensions: vertical (y-direction) and horizontal (x-direction). The vertical motion is influenced by gravity, while the horizontal motion is determined by the speed of the aircraft. ### Step 3: Determine the vertical motion The vertical distance (h) the projectile falls is given as 730 m. The time (t) taken to fall is 5 seconds. We can use the following equation of motion for vertical displacement: \[ h = V_{y} t + \frac{1}{2} g t^2 \] Where: - \( V_{y} \) is the vertical component of the velocity of the aircraft at the time of release. - \( g \) is the acceleration due to gravity (approximately \( 10 \, \text{m/s}^2 \)). ### Step 4: Calculate the vertical component of velocity The vertical component of the velocity \( V_{y} \) can be expressed as: \[ V_{y} = V \cos(53^\circ) \] Where \( V \) is the speed of the aircraft. Substituting this into the equation of motion gives: \[ 730 = V \cos(53^\circ) \cdot 5 + \frac{1}{2} \cdot 10 \cdot (5^2) \] ### Step 5: Substitute known values Calculating the second term: \[ \frac{1}{2} \cdot 10 \cdot 25 = 125 \] Now substituting \( \cos(53^\circ) \) (which is approximately \( \frac{3}{5} \)) into the equation: \[ 730 = V \cdot \frac{3}{5} \cdot 5 + 125 \] This simplifies to: \[ 730 = 3V + 125 \] ### Step 6: Solve for V Rearranging the equation gives: \[ 3V = 730 - 125 \] \[ 3V = 605 \] \[ V = \frac{605}{3} \approx 201.67 \, \text{m/s} \] ### Step 7: Conclusion The speed of the aircraft is approximately \( 202 \, \text{m/s} \).
To find the speed of the aircraft, we can break down the problem step by step. ### Step 1: Understand the problem We have an airplane diving at an angle of 53 degrees with the vertical. It releases a projectile from an altitude of 730 m, and the projectile hits the ground after 5 seconds. We need to find the speed of the aircraft. ### Step 2: Identify the components of motion The motion of the projectile can be analyzed in two dimensions: vertical (y-direction) and horizontal (x-direction). The vertical motion is influenced by gravity, while the horizontal motion is determined by the speed of the aircraft. ...
Promotional Banner

Similar Questions

Explore conceptually related problems

An airplane, diving at an angle of 5.3 0 ^@ with the vertical releases a projectile at an altitude of 730 m. The projectile hits the ground 5.00 s after being released. The speed of the aircraft is

A defense air force plane, diving with constant speed at an angle of 52.0^(@) with the vertical, drops a shell at an altitude of 720 m. The shell reaches the ground 6.00 s after its release. (a) What is the speed of the plane? (b) How far does the shell travel horizontally during its flight? What are the ( c) horizontal and (d) vertical components of its velocity just before reaching the ground? Assume an x axis in the direction of the horizontal motion and an upward y axis.

A bomber flying upward at an angle of 53^(@) with the vertical releases a bomb at an altitude of 800 m .The bomb strikes the ground 20 s after its release.If g=10ms^(-2) , the velocity at the time of release of the bomb in ms^(-1) is

A Bomber flying upward at an angle of 53^(@) with the vertical releases a bomb at an altitude of 800 m. The bomb strikes the ground 20 s after its release. Find : [Given sin 53^(@)=0.8, g=10 m//s^(2) ] (i) The velocity of the bomber at the time of release of the bomb. (ii) The maximum height attained by the bomb. (iii) The horizontal distance tranvelled by the bomb before it strikes the ground (iv) The velocity (magnitude & direction) of the bomb just when it strikes the ground.

The highest barrier that a projectile can clear is 13.5 m, when the projectile is launched at an angle of 15.0^(@) above the horizontal. What is the projectile's launch speed?

The range of a projectile, when launched at an angle of 15^(@) with the horizontal is 1.5 km. what is the range of the projectile, when launched at an angle of 45^(@) to the horizontal with the same speed ?

Particle A is projected with speed u at an angle theta with the horizontal . Another particle B is also projected with same speed u at an angle 90-theta with the horizontal . Both the projectiles are fired from the ground and reaches the ground after completing the projectile . Time of flight for A is t_(A) and that for B it is t_(B) . Range of the projectile A is R_(A) and that for B is R_(B) . Maximum height reached by the projectile A is h_(A) and that for B is h_(B)

An airplane with a speed of 97.5 m/s is climbing upward at an angle of 50.0^(@) with respect to the horizontal. When the plane's altitude is 732 m, the pilot releases a package, (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.