A ball is thrown horizontally from the top of a building 43.4 m high. The ball strikes the ground at a point 57.3 m from the base of the building.
The acceleration of gravity is 9.8 m/s^2.
a. Find the time the ball is in motion.
Answer in units of s.
b. Find the initial velocity of the ball. Answer in units of m/s.
c. Find the x component of its velocity just before it strikes the ground. Answer in units of m/s.
d. Find the y component of its velocity just before it strikes the ground. Answer in units of m/s.
a. h = 0.5g*t^2
h = 43.4 m.
g = 9.8 m/s^2
t = Fall time.
Solve for t.
b. Dx = Xo * t
Dx = 57.3 m. = Hor. dist.
Xo = Initial hor velocity.
t = Fall time calculated in part a.
Solve for Xo.
c. X = Xo=Initial hor velocity(part b).
Xo does not change.
d. Y = Yo + g*t
Yo = 0
g = 9.8 m/s^2
t = Fall time calculated in part a.
Solve for Y.
a. To find the time the ball is in motion, we can use the equation of motion for vertical motion:
h = (1/2)gt²
Where:
h = height (43.4 m)
g = acceleration due to gravity (9.8 m/s²)
Rearranging the equation to solve for time (t):
t = √(2h/g)
Plugging in the values:
t = √(2 * 43.4 / 9.8) ≈ √(88.8 / 9.8)
t ≈ √9.067 ≈ 3.01 s
Therefore, the time the ball is in motion is approximately 3.01 seconds.
b. To find the initial velocity (u) of the ball, we can use the equation:
u = (d / t)
Where:
d = horizontal distance (57.3 m)
t = time (3.01 s)
Plugging in the values:
u = 57.3 / 3.01 ≈ 19.03 m/s
Therefore, the initial velocity of the ball is approximately 19.03 m/s.
c. The x-component of its velocity just before it strikes the ground will remain the same throughout its motion. Therefore, the x-component of its velocity just before it strikes the ground will also be approximately 19.03 m/s.
d. The y-component of its velocity just before it strikes the ground can be found using the equation:
v = u + gt
Where:
u = initial velocity (19.03 m/s)
g = acceleration due to gravity (9.8 m/s²)
t = time (3.01 s)
Plugging in the values:
v = 19.03 + 9.8 * 3.01 ≈ 19.03 + 29.498
v ≈ 48.53 m/s
Therefore, the y-component of its velocity just before it strikes the ground is approximately 48.53 m/s.
a. To find the time the ball is in motion, we can use the formula for the time of flight of a projectile:
Time = distance / velocity
In this case, the distance the ball travels horizontally is 57.3 m. Since the ball is thrown horizontally, its initial vertical velocity is 0 m/s. Therefore, we do not need to consider the vertical component of the motion.
So, we just need to find the horizontal velocity of the ball. Since the ball is thrown horizontally, the horizontal velocity remains constant throughout the motion.
The horizontal velocity can be found using the equation:
velocity = distance / time
Therefore, the time the ball is in motion is:
Time = distance / velocity = 57.3 m / velocity
b. To find the initial velocity of the ball, we can use the formula for horizontal velocity:
velocity = distance / time
Since the ball is thrown horizontally, the initial vertical velocity is 0 m/s. Therefore, the initial velocity of the ball is equal to its horizontal velocity.
c. The x component of the velocity remains constant throughout the motion, and its value is equal to the initial horizontal velocity found in part b.
d. The y component of the velocity just before the ball strikes the ground can be found using the equation:
velocity_final = velocity_initial + (acceleration)(time)
Where:
velocity_final is the final vertical velocity of the ball just before it strikes the ground (y component of the velocity).
velocity_initial is the initial vertical velocity of the ball (0 m/s in this case, since it's thrown horizontally).
acceleration is the acceleration due to gravity (9.8 m/s^2).
time is the time of flight of the ball calculated in part a.
So, the y component of the velocity just before the ball strikes the ground is:
velocity_final = 0 m/s + (9.8 m/s^2)(time)
To solve this problem, we can use the equations of motion for an object in freefall.
Let's start by identifying what we know:
Initially:
- The initial height (h) of the ball is 43.4 m.
- The initial vertical velocity (Vy) is 0 m/s, as the ball is thrown horizontally.
At the end:
- The final height (hf) is 0 m, as the ball strikes the ground.
- The horizontal displacement (x) is 57.3 m.
To solve part (a):
The time the ball is in motion can be found using the kinematic equation:
hf = hi + Vyi * t + (1/2) * g * t^2,
where hi is the initial height, Vyi is the initial vertical velocity, g is the acceleration due to gravity, and t is time.
Since the initial vertical velocity is 0, the equation simplifies to:
hf = hi + (1/2) * g * t^2.
Plugging in the known values:
0 = 43.4 + (1/2) * 9.8 * t^2.
Rearranging the equation to isolate t:
0 = 43.4 + 4.9 * t^2.
4.9 * t^2 = -43.4.
t^2 = -43.4 / 4.9.
t^2 = -8.857.
Since time cannot be negative, there is no real solution for t. Hence, the ball is not in motion.
To solve parts (b), (c), and (d):
Since the ball is not in motion, there is no vertical or horizontal velocity to determine. Therefore, the answers for parts (b), (c), and (d) are not applicable in this case.