Question

25. A stone is thrown vertically upward with a speed of
28 m/s(a) Find the maximum height reached by the
stone. (b) Find its velocity one second before it reaches
the maximum height. (C) Does the answer of part
(b) change if the initial speed is more than 28 m/s such
as 40 m/s or 80 m/s ?
​

Answer 1

Answer:

Explanation:

a) $h=u^2/2g\\h=(28)^/19.6\\h=784/19.6\\h=40m$

b) Deceleration is 9.8 m/s². Which means every second 9.8 m/s velocity is reduced. So one second before reaching top, v= 9.8m/s. This can also be derived as:

$v=u-gt\\0=u-g(1)\\u_{last.sec}=g$

c) Look at above derivation for $u__{last.sec}$, it doesn't depend on initial velocity of projectile.

Answer 2

Given :-

Speed of stone is thrown vertically upward = 28 m/s

To Find :-

The maximum height reached by the  stone.

It's velocity one second before it reaches the maximum height.

Does the answer of part change if the initial speed is more than 28 m/s such as 40 m/s or 80 m/s

Solution :-

We know that,

• v = Final velocity
• u = Initial velocity
• t = Time
• a = Acceleration

Given that,

A stone is thrown vertically upward with a speed of 28 m/s.

$\implies \sf u = 28 \ m/s$

Final velocity (v) = 0 (when stone reaches the ground)

Acceleration due to gravity (a) = g = $\sf 9.8 \ m/s^{2}$

Solving (a):

We know that,

$\sf V^{2} - u^{2} = 2as$

Or s = 40 m

The maximum height reached by the stone is 40 m.

Solving (b):

Using the relation, V = u + at

$\sf Or \ t =\dfrac{(v-u)}{a}$

$\sf Or \ t = 2.85 \ s$

According to the question,

We need to find the velocity of the stone one second before it reaches the maximum height.

$\sf t ' = 2.85 - 1 = 1.85 \ s$

Now again,

$\sf v' = u + at' = 28 - 9.8 \times 1.85$

$\sf = 9.87 \ m/s$

The required velocity is 9.87 m/s

Solving (c):

No, the answer part doesn't change if the initial speed is more than 28 m/s such  as 40 m/s or 80 m/s

After one second, the velocity becomes zero for any initial velocity and acceleration $\sf (a = - 9.8 \ m/s^{2})$ remains the same.