Question

You throw a ball vertically upward from a height of 2 meters with an initial velocity of 10 meters per second. Use the idea that acceleration due to gravity (neglecting air resistance) is -32 feet/sec^2, which is the same as -9.8 meters/sec^2.

a. What is the maximum height the ball will reach? Round your answer jot the nearest meter, if necessary.

b. How many seconds will go by before the ball hits the ground after being thrown? Round your answer to the nearest tenth of a second, if necessary.

Answer 1

Here height, velocity and time are represented by x, v and t respectively.

Initial height= 2 m

Initial velocity = 10 m/s

As the ball thrown vertically upward acceleration is -9.8 m/s²

(a) $\frac{dv}{dt}$ = -9.8

=> $\frac{dv}{dx} \frac{dx}{dt}$ = -9.8

=> v$\frac{dv}{dx}$ = -9.8 as $\frac{dx}{dt}$ = v

=> vdv = -9.8dx

=> $\int$ vdv = -9.8 $\int$ dx

=> $\frac{v^2}{2}$ = -9.8x + C , C is integration constant

By initial condition , when x = 2, v = 10

So $\frac{100}{2}$ = -19.6 + C => C = 69.6

Therefore $\frac{v^2}{2}$ = -9.8x + 69.6

=> v² = -19.6x + 139.2

The ball will gain height as it will have a positive velocity.

So the height will be maximum when v= 0

Now putting v= 0 we get

-19.6x + 139.2 = 0

=> x = $\frac{139.2}{19.6}$ = 7 ( nearest to integer)

So maximum height the ball will reach is 7 m ( nearest to meter)

(b) To solve this part we will find the relation between height (x) and time (t)

Now $\frac{dv}{dt}$ = -9.8

=> dv = -9.8dt

=> $\int$ dv = -9.8 $\int$ dt

=> v = -9.8t + C₁ where C₁ is integration constant.

By initial condition , when t= 0, v = 10

So C₁ = 10

Therefore v = 10 - 9.8t

=> $\frac{dx}{dt}$ = 10 - 9.8t

=> dx = 10dt - 9.8tdt

Integrating

$\int$ dx = 10 $\int$ dt - 9.8$\int$ tdt

=> x = 10t - 9.8 $\frac{t^2}{2}$ + C_2. where C₂ is integration constant.

When t = 0, x = 2

So 2 = C₂

Therefore x = 2 + 10t - 4.9t²

When the ball will reach the ground, height of the ball, x = 0

So putting x= 0 we get

2 + 10t - 4.9t² = 0

=> 4.9t² - 10t -2 = 0

=> 49t² - 100t - 20 = 0

Using quadratic formula

t = $\frac{100±\sqrt{100²+4*49*20}}{2*49}$ = $\frac{100±\sqrt{13920}}{98}$

So t = $\frac{100+117.983}{98} or \frac{100-117.983}{98}$

=> t = 2.224 or - 0.1835

Negative value of t is rejected as time can not be negative.

So t = 2.224 = 2.2 ( correct upto one place of decimal)

So time taken to hit the ground is 2.2 second