A bomber plane moves horizontally with a speed of 500 m/s and a bomb released from it, strikes the ground in 10 s. Angle at which it strikes the ground will be: (g = 10m/s2)

To solve this problem, we need to find the horizontal and vertical distances travelled by the bomb.

Step 1: Calculate the horizontal distance (d) travelled by the bomb. The horizontal speed of the bomb is the same as the speed of the plane, which is 500 m/s. Since the bomb hits the ground in 10 seconds, the horizontal distance travelled by the bomb is: d = speed * time d = 500 m/s * 10 s = 5000 m

Step 2: Calculate the vertical distance (h) travelled by the bomb. The bomb is dropped from the plane, so it falls to the ground due to gravity. The distance an object falls due to gravity can be calculated using the formula: h = 0.5 * g * t^2 where g is the acceleration due to gravity (10 m/s^2) and t is the time the bomb is falling (10 s). So, h = 0.5 * 10 m/s^2 * (10 s)^2 = 500 m

Step 3: Calculate the angle (θ) at which the bomb hits the ground. The angle can be calculated using the tangent of the angle (tan θ), which is the vertical distance divided by the horizontal distance. So, tan θ = h / d tan θ = 500 m / 5000 m = 0.1 To find the angle, we take the inverse tangent (arctan) of 0.1, which gives us: θ = arctan(0.1) = 5.71 degrees

So, the bomb hits the ground at an angle of approximately 5.71 degrees.