Question

A pebble is thrown vertically upward from a
bridge with an initial velocity of 4.9 ms. It strikes the water
after 2 s. If acceleration due to gravity is 9.8 ms" (i) what is
the height of the bridge? (ii) with what velocity does the pebble
strike the water?​

Answer 1

Answer:

1. The height of the bridge is 9.8 m.
2. The velocity with which  the pebble strike the water surface is 14.7 ms⁻¹

Explanation:

Given :

• Initial velocity - 4.9 ms.
• Time - 2 seconds.
• Gravity - 9.8 ms.

To Find :

• The height of the bridge.
• At which velocity does the pebble strike the water.

Solution :

1 : The height of the bridge.

Consider the :

The height of the bridge above water surface as - h.

Now,

Time t₁ taken to reach the highest point from bridge is given by :

$\bigstar\;\boxed{\bf v=u+at}}$

$\sf\\ \\\implies 0=4.9-9.8\;t_1\\ \\\implies t_1 = 0.5\;sec$

Now,

The height attained is given by :

$\sf\\ \\\implies v^2=u^2+2as\\ \\\implies 0=(4.9)^2+2(-9.8)\;h_1\\ \\\implies h_1 = \dfrac{4.9\times4.9}{2\times9.8} = 1.225m$

Since the total time for journey is 2 sec,

∴ The total time for total downward journey is :

⇒ t = 2 - 0.5 sec

⇒ 1.5 sec

From the relationship,

$\sf\\ \\\implies S=ut+\dfrac{1}{2}\;at^2\\ \\\implies (h+h_1)=0+\dfrac{1}{2}\times9.8\times(1.5)^2\\ \\\implies h=4.9 \times2.25-1.225\\ \\\implies 9.8m$

∴ The height of the bridge is 9.8 m.

$\rule{300}{1.5}$

2 : At which velocity does the pebble  strike the water.

$\bigstar\;\boxed{\bf v=u+at}}$

$\sf\\ \\\implies 0+9.8\times1.5\\ \\\implies 14.7\;ms^{-1}$

∴ The velocity with which  the pebble strike the water surface is 14.7 ms⁻¹