Question

A particle is thrown up vertically with a velocity of 75m/s. (a) What will be its velocity at the highest point of its journey? (b) How high would the particle rise? (c) What time would it take to reach the highest point?

Answer 1

Answer:

(a) zero

(b) 287m

(c) 7.65 seconds.

Explanation:

Given,

A particle is thrown up vertically with a velocity of 75m/s.

Here, after reaching a certain height the particle will come to stop due to gravity.

Hence, velocity at its highest point will be zero.

$\therefore \: v \rm \: = 0m \: {s}^{ - 1}$

We have,

• $u\rm(initial \: velocity) = 75m\: s^{-1}$
• $g\rm(acceleration \: due \: to \: gravity) = -9.8m \: s^{-1}$

Note:- [Negative sign denotes the opposition of gravitational force.]

The highest point it would rise is given by,

$\implies 2gs = {v}^{2} - {u}^{2}$

Here,

• $s$ denotes the distance.

$\implies 2( - 9.8)s = {(0)}^{2} - {(75)}^{2}$

$\implies - 19.6s = 0 - 5625$

$\implies s = \dfrac{ - 5625}{ - 19.6}$

$\implies s = 286.98$

$\therefore \boldsymbol s \approx 287$

∴ The particle would rise a height of 287m.

The time taken is given by,

$\implies v = u + gt$

Substitute the values,

$\implies 0 = 75 + ( - 9.8)t$

$\implies - 75 = - 9.8t$

$\implies \dfrac{ - 75}{ - 9.8} = t$

$\therefore \boldsymbol t = 7.65$

∴ Time taken to reach the height is 7.65 seconds.

__________________________

Answer 2

Answer:

(a) 0

(b) 287 m

(c) 7.65 seconds

Explanation:

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