Question

Harry and Indian fighter plane flying horizontally with speed 800 metre per hour release a bomb on Pakistani Bunker at a height of 78.4 m from the ground when will bomb strike the ground
(g= 9.8m/s^2)

Answer 1

First convert speed 800 km/h in x m/s

800×1000/3600=222.22m/s

When the bomb is dropped ,it will have an initial horizontal velocity which is equal to the speed of the plane. So the bomb fall and will travel forward.

Initial horizontal velocity(v)=222.22m/s

Vertical velocity(u)=0m/s

Height from which bomb is dropped(h)=78.4m

Time=tsecs

Now from 2nd equation of motion h=ut+1/2gt²

78.4= 0(t)+(1/2)×9.8×t²

78.4= 4.9t²

t²= 78.4/4.9= 16

t= 4 sec

So 4secs is the total time which bomb take to reach ground by travelling y (let) distance.

So y= vt = 222.22m/s × 4secs = 888.88m

Distance travel by bomb =888.88m

Answer 2

We can solve this problem using the equations of motion.

Let's first find the time taken by the bomb to reach the ground.

Using the equation of motion for vertical motion, we get:

h = ut + 1/2 * gt^2

Substituting the given values, we get:

78.4 = 0 + 1/2 * 9.8 * t^2

Simplifying this equation, we get:

t^2 = 16

t = 4 seconds (taking the positive square root as time cannot be negative)

So the bomb will take 4 seconds to reach the ground.

Now, let's find the horizontal distance traveled by the bomb during this time. We know that the horizontal velocity of the bomb is 800 m/hour, which is the same as 800/3600 = 0.2222 m/s.

Using the equation of motion for horizontal motion, we get:

s = vt

Substituting the given values, we get:

s = 0.2222 * 4

s = 0.8888 meters

Therefore, the bomb will strike the ground at a horizontal distance of 0.8888 meters from the point directly below the point where it was released.

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