A volcano shoots old blobs of molten lava called lava bombes from its summit. A geologist observing the irruption uses a stopwatch to time the flight of a particular lava bomb that is projected straight upwards. If the time for the bomb to rise and fall back to its launch height is 4.75 s, and it's acceleration is 9.8 m/s^2 downward, what is its initial speed?

The initial speed of bomb is 46.55 m/s.

Given data:

The time required to rise and fall is, t = 4.75 s.

The magnitude of gravitational acceleration is, .

The given problem is based on the free fall motion. And we can use the first kinematic equations of motion to obtain the value of initial speed (u) as,

v = u + gt

here,

v is the final speed of bomb. And its value is, v = 0 ( Because after the maximum height, the bomb will stop).

Solving as,

0 = u + (-9.8)(4.75)

u = 46.55 m/s

Thus, we can conclude that the initial speed of bomb is 46.55 m/s.

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46.55m/s

Explanation:

Given

Acceleration of the body = 9.8 m/s^2

time for the bomb to rise and fall back to its launch height = 4.75 secs

final speed = 0m/s

Required

Initial speed

Using the equation of motion

v = u + gt

Since the bomb is project upwards the acceleration due to gravity will be negative, hence;

v = u -gt

Substitute the given values

0 = u - 9.8(4.75)

-u = -9.8(4.75)

u = 9.8(4.75)

u = 46.55m/s

Hence its initial speed is 46.55m/s

c. magnitude and direction