Question

A ball with mass m kg is thrown upward with initial velocity 22 m/s from the roof of a building 25 m high. Neglect air resistance. Use g=9.8 m/s2. Round your answers to one decimal place. (a) Find the maximum height above the ground that the ball reaches. xmax= meters (b) Assuming that the ball misses the building on the way down, find the time that it hits the ground. tend= s

Answer 1

Answer:

(a) 24.7 metres

(b) 5.4 seconds

Explanation:

(a) To solve this part, we only need the initial velocity of the ball, u = 22 m/s

We apply one of Newton's equations of motion to solve this:

v² = u² - 2gs

Where v = final velocity

u = initial velocity

g = acceleration due to gravity = 9.8 m/s

s = distance moved by ball

Note: The equation has a negative sign because the ball is moving against the gravitational force.

The maximum height reached by the ball will be attained when the ball's final velocity, v, is 0 m/s.

Hence:

0² = 22² - 2 * 9.8 * s

=> 19.6s = 22²

19.6s = 484

s = 484 / 19.6 = 24.7 m

The maximum height reached by the ball is 24.7 m

(b) We have to solve two parts, the time it take a takes the ball to reach its maximum height and the time it takes to fall after reaching maximum height.

TIME IT TAKES TO REACH MAXIMUM HEIGHT

Using another one of Newton's equations, we have that:

v = u - gt

We already have that v = 0 m/s and u = 22 m/s.

Hence:

0 = 22 - 9.8t

9.8t = 22

=> t = 22 /9.8 = 2.2 seconds

TIME IT TAKES TO FALL AFTER REACHING MAXIMUM HEIGHT

Since the ball was thrown from a height of 25 m and it has now reached 24.7 m higher, the ball is now at a height of:

25 + 24.7 = 49.7 m

We apply another one of Newton's equations of motion to obtain the time it will take to reach the ground:

s = ut + ½gt²

t = time taken

The initial velocity will become zero (0 m/s) at the point because the ball is beginning its descent.

Hence:

49.7 = (0*t) + ½ * 9.8 * t²

49.7 = 4.9t²

=> t² = 49.7 / 4.9 = 10.14

=> t = 3.2 seconds

It will take the ball 3.2 seconds to hit the ground after it starts descending.

The total time it takes the ball to hit the ground will therefore be:

T = 2.2 + 3.2 = 5.4 seconds

Answer 2

Answer:

A) x_max = 56.4 m

B) the time that the ball hits the ground from maximum height = 5.425 seconds

Explanation:

A) The differential equation of the motion is given by;

m(dv/dt) = -mg

Dividing each side by m gives;

dv/dt = -g

Thus, dv = -g•dt

Taking the integral of both sides gives;

∫dv = -∫g•dt

This gives;

v = -gt + c

From, the question, we are given that initial velocity is 22 m/s i.e, c=22.

Thus at max height, v = 0, so;

0 = -gt + 22

So,making t the subject gives;

t = 22/g

g is acceleration due to gravity = 9.81 m/s²

t = 22/9.81

t = 2.243 s

Now, the position equation is calculated from;

x(t) = ∫v(t)•dt = ∫(-gt + 22)•dt

= -½gt² + 22t + c

Initial position from the question is 25m, that's c = 25,thus;

x(t) = -½gt² + 22t + 25

Plugging in the value of t = 2.243 and g = 9.81 gives;

x(t) = -½*9.81*2.243² + 22(2.243) + 25

x(t) = 56.4m

B) The ball hits the ground when x(t) = 0.

Thus; since the equation of x(t) is;

x(t) = -½gt² + 22t + 25

We now have;

-½gt² + 22t + 25 = 0

Putting in 9.81 for g gives;

-4.905t² + 22t + 25 = 0

Finding the roots using quadratic formula, gives;

t = 5.425 or -0.94

We'll ignore the negative value and pick the positive one as time cannot be negative.

Thus;

t = 5.425 seconds