Question

A rock is thrown straight up and reaches a height of 10 m
a) how long was the rock in the air
b) what is the initial velocity of the rock

Answer 1

Given, h${\tinymax}$ = 10 m

final velocity, v = 0 m/s

acceleration due to gravity, g = -9.8 m/s^2(in upward direction)

a)

Using motion first equation,

2gh = v^2 - u^2

=> 2 × (-9.8) × 10 = 0 - u^2

=> (-196) = (-u^2)

=> 196 = u^2

=> u = √196 = 14 m/s

time of flight, t = $\frac{2 u}{g}$

=$\frac{2×14}{10}$

= 2.8s (approx.)

b)

initial velocity, u = 14 m/s

Answer 2

Answer:

The rock is in the air for 2.856 s and initial velocity of the rock is 14 m/s.

Explanation:

As the rock is thrown upwards, it experiences the acceleration due to gravity downwards. The final velocity of the rock is zero as at the top rock stops.

It is given that the displacement performed by the rock is the height it reaches, s = 10 m

The acceleration due to gravity is given by $g = - 9.8 m/s^{2}$

Acceleration due to gravity is negative because it acts in downward direction.

The final velocity of the rock, v = 0 m/s

Let the initial velocity be u.

So, according to the third equation of motion,

$v^{2} - u^{2} = 2gs$ (1)

Putting the values in (1) equation,

$0 - u^{2} = 2(-9.8)(10)$

$u^{2} = 196$

$u = 14 m/s$

So, the initial velocity of rock is 14 m/s.

Now, the initial velocity of rock, u = 14 m/s

The final velocity of rock, v = 0 m/s

Acceleration due to gravity, g = -9.8 m/s

Let the time be t.

So, according to the first equation of motion,

$v = u + at$ (2)

Putting the values in (2) equation,

$0 = 14 + (-9.8)(t)$

$t = 140/98$

$t = 1.428 s$

The rock also come to initial position. So, double of time t is the time taken to be in air.

So, the time taken for the rock to be in air = 2×1.428

= 2.856 s