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During volcanic eruptions, chunks of solid rocks can be blasted out of the volcano; these projectiles are called volcanic blocks. If a volcano is 3.5km high

and the base of the volcano is 5.6 km from the centre of the volcano and a block is ejected at 317m/s at an angle of 43 to the horizontal.
What is the time of flight for the volcanic block assuming it falls the full height of the volcano?
a) 5.2 s
b) 10.4 s
c) 13.2 s
d) 26.4 s

User Prabhat
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1 Answer

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Final answer:

To calculate the time of flight for the volcanic block, we use the initial vertical velocity and the height of the volcano, considering the acceleration due to gravity. The block's vertical motion determines the total time of flight, which includes the ascent to the maximum height and the subsequent descent to the base of the volcano.

Step-by-step explanation:

To find the time of flight for a volcanic block ejected from a volcano, we need to consider the vertical motion of the block under the influence of gravity. The block was projected upward at an angle of 43 degrees relative to the horizontal with an initial velocity of 317 m/s.

We can decompose this velocity into its horizontal and vertical components with trigonometric functions, but we only need the vertical component (initial vertical velocity) to find the time of flight.

Using the formula for the vertical motion y = Vot + 0.5at^2, where y is the displacement, Vo is the initial vertical velocity, t is time, and a is the acceleration due to gravity (which is approximately -9.81 m/s^2 since it's acting downwards), we can solve for t when y is equal to -3.5 km (the height of the volcano that the block falls).

Since we are looking for the time of flight, we are interested in the total time it takes for the block to reach its peak and then fall down to the base of the volcano. This time is obtained by doubling the time it takes to reach the peak, as the ascent and descent times are equal in the absence of air resistance.

User Innocent
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