Question

A ball is thrown straight up into the air with a speed of 21 m/s. If the ball has a mass of 0.1 kg, how high does the ball go? Acceleration due to gravity is g = 9.8 m/s^2.
A. 22.5 m
B. 20.0 m
C. 25.0 m
D. 17.5 m

Answer 1

Hi!

The answer would be 22.5m

Explanation

To calculate the distance the ball travels, you would need to apply the following formula from the equations of motion for an accelerating object:

2as = v^2 - u^2

Where a is acceleration due to gravity, 9.8m/s^2

s is the distance traveled by the object (height of ball in this case)

v is the final velocity, which we know will be zero at the point where the ball reaches maximum height.

u is the initial velocity, which is known to us as 21m/s

Rearranging the equation to solve for the height:

s = (v^2 - u^2 ) / 2a

s = ( 0^2 - 21^2 ) / 2(-9.8)

s = - 441/ - 19.6

s = 22.5m

Note: since gravity is acting against the object's motion, it will be negative

Hope this helps!

Answer 2

Correct answer is A(22.5m)

In this question we have given,

Initial velocity, $u=21\frac{m}{s}$

Final velocity, at maximum height, $v=0$

mass of ball, m=.1kg

Acceleration due to gravity is g = 9.8$\frac{m}{s^2}$

we have to find the maximum height attained by the ball,h=?

We know by third equation of motion,

$v^2=u^2-2as$...................(1)

Here,a=g=9.8$\frac{m}{s^2}$

and s=h

Put values of v,u,a and s in equation(1)

$0=21^2\frac{m^2}{s^2}-2\times 9.8\frac{m}{s^2}h$

$0=441\frac{m^2}{s^2}-19.6\frac{m}{s^2}h$

$-441\frac{m^2}{s^2}=-19.6\frac{m}{s^2}h$

$-441\frac{m^2s^2}{-19.6ms^2}=h$

or h=22.5m

therefore, the maximum height attained by the ball,h=22.5m