a ball is thrown up with an initial velocity 20m/s. using g=10m/s2,
a) what's the maximum height it can reach.
b) what is the magnitude and direction of the ball's velocity 1 second after it is thrown.
c) what is the magnitude and direction of the ball's velocity 3 seconds after it is thrown
a. h = (Vf^2 - Vo^2) / 2g,
h = (0 - (20)^2) / 20 = 20m.
b. Vf = Vo + gt,
Vf = 20 + (-10)*1 = 10m/s,up.
c. Vf = 20 + (-10)*3 = -10m/s,down.
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a) To calculate the maximum height reached by the ball, we can use the equation for vertically thrown objects:
h = (v^2 / 2g)
where:
h = maximum height
v = initial velocity
g = acceleration due to gravity
Plugging in the values:
v = 20 m/s
g = 10 m/s^2
h = (20^2 / 2 * 10)
h = (400 / 20)
h = 20 meters
So, the maximum height reached by the ball is 20 meters.
b) To calculate the magnitude and direction of the ball's velocity after 1 second, we need to consider the acceleration due to gravity.
The magnitude of the velocity can be calculated using the equation:
v = u + at
where:
v = final velocity
u = initial velocity
a = acceleration
t = time
Plugging in the values:
u = 20 m/s
a = -10 m/s^2 (negative due to the ball moving upward)
t = 1 second
v = 20 + (-10 * 1)
v = 20 - 10
v = 10 m/s
So, the magnitude of the ball's velocity after 1 second is 10 m/s. Since the ball is moving upwards, the direction of the velocity is upwards.
c) To calculate the magnitude and direction of the ball's velocity after 3 seconds, we again use the same equation:
v = u + at
Plugging in the values:
u = 20 m/s
a = -10 m/s^2 (negative due to the ball moving upward)
t = 3 seconds
v = 20 + (-10 * 3)
v = 20 - 30
v = -10 m/s
So, the magnitude of the ball's velocity after 3 seconds is 10 m/s. Since the ball is moving downwards, the direction of the velocity is downwards.
To find the answers to these questions, we can use the equations of motion for an object thrown vertically upwards.
a) To find the maximum height the ball can reach, we need to use the equation:
H = (V^2) / (2g)
Where:
H is the maximum height
V is the initial velocity
g is the acceleration due to gravity
Given:
V = 20 m/s
g = 10 m/s^2
Substituting the values into the equation, we get:
H = (20^2) / (2 * 10) = 400 / 20 = 20 meters
Therefore, the maximum height the ball can reach is 20 meters.
b) To determine the magnitude and direction of the ball's velocity 1 second after it is thrown, we use the equation:
V = U - gt
Where:
V is the final velocity
U is the initial velocity
g is the acceleration due to gravity
t is the time
Given:
U = 20 m/s
g = 10 m/s^2
t = 1 second
Substituting the values into the equation, we get:
V = 20 - (10 * 1) = 20 - 10 = 10 m/s
The magnitude of the ball's velocity after 1 second is 10 m/s. Since the ball is moving upward, the direction of the velocity is upwards.
c) To find the magnitude and direction of the ball's velocity 3 seconds after it is thrown, we can use the same equation as above:
V = U - gt
Given:
U = 20 m/s
g = 10 m/s^2
t = 3 seconds
Substituting the values into the equation, we get:
V = 20 - (10 * 3) = 20 - 30 = -10 m/s
The magnitude of the ball's velocity after 3 seconds is 10 m/s. However, since the ball is now moving downward, the direction of the velocity is downwards. The negative sign indicates the change in direction from upwards to downwards.