A ball is thrown horizontally from a building 50.0 m high with a speed of 18 m/s. (a) find the vertical and horizontal components of the ball's initial velocity, (b)time of flight, (c) distance from the foot of the building where the ball will strike the ground, and (d) velocity when the ball will strike the ground.
You threw it horizontally so the initial vertical velocity is zero and the horizontal component is 18
Divide into a vertical problem and a horizontal problem
Verttical (falling) problem
Vi = 0
Hi = 50
a = -g = -9.81 meters/ second^2
v = 0 - 9.81 t
h = 50 - 0 t - 4.9 t^2
at bottom h = 0
4.9 t^2 = 50
t = 3.2 second falling time to ground
v = -9.81* 3.2 = - 31.3 meters/ second at h = 0, ground
HORIZONTAL problem
the thing flies at 18 m/s for 3.2 seconds
x = 18 * 3.2 meters
Adrian threw a ball vertically upward, it return to his hands after 15seconds.(A) what was the ball initial speed? (B)with what speed did it return to the adrian's hands? (C) How high did it go?.
Can anyone answer it in complete solution thank you in advance
To solve this problem, we need to use the equations of motion for projectile motion. Projectile motion refers to the motion of an object that is launched into the air and moves under the influence of gravity alone. In this case, the ball is thrown horizontally, so its initial vertical velocity is 0 m/s.
(a) To find the horizontal and vertical components of the ball's initial velocity, we can use the following equations:
Horizontal component of velocity (Vx) = Initial Velocity (V0) * cos(angle)
Vertical component of velocity (Vy) = Initial Velocity (V0) * sin(angle)
Given that the initial velocity (V0) = 18 m/s and the angle of projection is 0° (horizontal), the horizontal and vertical components of velocity can be calculated as follows:
Vx = 18 m/s * cos(0°) = 18 m/s
Vy = 18 m/s * sin(0°) = 0 m/s
Therefore, the horizontal component of the ball's initial velocity is 18 m/s, and the vertical component is 0 m/s.
(b) Time of flight is the total time it takes for the ball to reach the ground. Since the vertical component of the initial velocity is 0 m/s, we can use the following equation for the vertical motion:
Vertical displacement (d) = Vertical component of initial velocity (Vy) * Time (t) + (1/2) * Acceleration due to gravity (g) * Time (t)^2
Given that the vertical displacement (d) is equal to -50.0 m (negative because it's moving downward, opposite to the positive y-axis), and the acceleration due to gravity (g) is approximately 9.8 m/s^2, we can rearrange the equation to find the time of flight (t):
-50.0 m = 0.5 * 9.8 m/s^2 * t^2
Solving for t, we get:
t^2 = (2 * -50.0 m) / 9.8 m/s^2
t^2 = -10.2
(The negative sign indicates that we should consider the positive root, as time cannot be negative)
t ≈ √10.2 ≈ 3.19 s
Therefore, the time of flight is approximately 3.19 s.
(c) The distance from the foot of the building where the ball will strike the ground can be found using the following equation for horizontal motion:
Horizontal displacement (d) = Horizontal component of initial velocity (Vx) * Time (t)
Given that the horizontal displacement (d) is what we need to find and the time of flight (t) is approximately 3.19 s, the equation becomes:
d = 18 m/s * 3.19 s = 57.42 m
Therefore, the ball will strike the ground approximately 57.42 m away from the foot of the building.
(d) The velocity when the ball strikes the ground can be found by considering the vertical component only, as the horizontal component remains constant throughout the motion. Since the ball is dropped from the maximum height, the final vertical velocity (Vfy) when it hits the ground can be found using the equation:
Vertical component of final velocity (Vfy) = Vertical component of initial velocity (Vy) + Acceleration due to gravity (g) * Time (t)
Given that the vertical component of initial velocity (Vy) is 0 m/s, the acceleration due to gravity (g) is approximately 9.8 m/s^2 and the time of flight (t) is approximately 3.19 s, the equation becomes:
Vfy = 0 m/s + 9.8 m/s^2 * 3.19 s
Vfy ≈ 31.262 m/s
Therefore, the velocity when the ball strikes the ground is approximately 31.262 m/s.