A ball is thrown horizontally from the top of a building 45.9 m high. The ball strikes the ground at a point 121 m from the base of the building.
The acceleration of gravity is 9.8 m/s2 . Find the time the ball is in motion. Answer in units of s
Find the initial velocity of the ball. Answer in units of m/s
Find the x component of its velocity just be- fore it strikes the ground.
Answer in units of m/s
Find the y component of its velocity just be- fore it strikes the ground.
Answer in units of m/s
I don't get it
To find the time the ball is in motion, we can use the equation of motion for vertical motion:
h = (1/2)gt^2
where h is the initial height of the ball (45.9 m), g is the acceleration due to gravity (-9.8 m/s^2), and t is the time. Since the ball starts from rest horizontally and falls vertically, the time it takes for the ball to reach the ground is the same as the time it is in motion.
Substituting the given values into the equation:
45.9 = (1/2)(-9.8)t^2
Simplifying the equation:
t^2 = (2 * 45.9) / 9.8
t^2 = d / g
t = sqrt(d / g)
t = sqrt(4.68)
t ≈ 2.163 seconds
Therefore, the time the ball is in motion is approximately 2.163 seconds.
To find the initial velocity of the ball, we can use the formula for horizontal motion:
d = vt
where d is the horizontal displacement (121 m), v is the initial velocity, and t is the time. Since the ball is thrown horizontally, its initial vertical velocity is 0.
Substituting the given values into the equation:
121 = v * 2.163
v = 121 / 2.163
v ≈ 56.0 m/s
Therefore, the initial velocity of the ball is approximately 56.0 m/s.
To find the x component of the velocity just before the ball strikes the ground, we can use the formula for horizontal motion:
v_x = v
where v_x is the horizontal component of the velocity and v is the initial velocity (56.0 m/s).
Therefore, the x component of the velocity just before the ball strikes the ground is 56.0 m/s.
To find the y component of the velocity just before the ball strikes the ground, we can use the equation of motion for vertical motion:
v_y = gt
where v_y is the vertical component of the velocity, g is the acceleration due to gravity (-9.8 m/s^2), and t is the time it takes for the ball to reach the ground (2.163 seconds).
Substituting the given values into the equation:
v_y = -9.8 * 2.163
v_y ≈ -21.2 m/s
Therefore, the y component of the velocity just before the ball strikes the ground is approximately -21.2 m/s.
Find the time to hit using
4.9t^2 = 45.9
Plug that t into get Vx
Vx = 121/t
Vx does not change
Vy = -9.8t