Respuesta :
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:
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
