Question

. A ball is thrown upward with an initial velocity of 4 ft/sec from a height of 10 feet. What is the maximum height that the ball will reach?

Answer 1

10.25 ft

Step-by-step explanation:

According to the law of conservation of energy, the initial kinetic energy of the ball when it is thrown upward is entirely converted into gravitational potential energy when the ball reaches its maximum height. So we can write:

`(1)/(2)mv^2 = mg\Delta h`

where

m is the mass of the ball

v = 4 ft/s is the initial speed

g = 32 ft/s^2 is the acceleration due to gravity

`\Delta h` is the change in height of the ball

Re-arranging the equation, we find

`\Delta h=(v^2)/(2g)=((4)^2)/(2(32))=0.25 ft`

And since the ball was thrown from a heigth of 10 feet, the maximum height reached is

h = 10 ft + 0.25 ft = 10.25 ft

Answer 2

10.248 ft

Step-by-step explanation:

Given that velocity u=4 ft/sec

Initial height=10 ft

We know that gravity is constant so we can apply

`v^2=u^2+2as`

Here we need to find maximum height .When it velocity will become zero then attained height will be maximum so v=0 , =4 ft/sec.

Now by putting the values

`0 = 16-2 * 32.17* h`

So h=0.248 ft

So the attained maximum height=0.248+10

=10.248 ft