Last two problems on the homework! please help with these 2. it helps when detailed explanations are given! thanks
A 2 kg ball is thrown horizontally, at 5m/s, from a height of 5 m. How fast will it be moving when it contacts the ground?
A 70 kg pole-vaulter converts the kinetic energy, of running at ground level, into potential energy to clear the crossbar at a height of 4 m above the gound. What is the minimum velocity that he must have had at takeoff (while on the ground) to clear the bar (minimum would be when velocity at the top was zero – you could still be moving forward at the top, but this would require more velocity at the beginning)?
u = 5 m/s forever
v = -g t
h = 5 - 4.9 t^2
h = 0 at ground
so
4.9 t^2 = 5
t = 1.01 seconds
so v = - 9.81 * 1.01 = - 9.91 m/s
speed = sqrt (5^2 + 9.91^2) = 11.1 m/s
in other words I dissagree with jzee11
ball's velocity 25 m/s when it hits the ground.
minimum velocity required is 8.85 m/s
how did you get those?
thank you so much!
You are welcome.
To solve these problems, we can use the principles of kinematics and energy conservation. Let's tackle each problem step by step.
Problem 1:
We have a 2 kg ball thrown horizontally from a height of 5 m with an initial velocity of 5 m/s. We want to find out how fast it will be moving when it contacts the ground.
To solve this problem, we need to analyze the motion of the ball in the vertical direction and horizontally.
1. In the vertical direction:
The ball experiences free fall due to gravity. We can use the equation for vertical displacement:
Δy = v_i*t + (1/2)*a*t^2
where Δy is the vertical displacement, v_i is the initial vertical velocity, t is time, and a is acceleration (which is equal to -9.8 m/s^2 for objects falling near the Earth's surface).
Given that the ball falls from a height of 5 m and the initial vertical velocity is 0 m/s (as it is thrown horizontally), we can write the equation as:
-5 = 0*t + (1/2)*(-9.8)*t^2
Simplifying:
-4.9t^2 = -5
t^2 = 5/4.9
t ≈ 1.02 seconds
So, the ball will take approximately 1.02 seconds to reach the ground.
2. In the horizontal direction:
Since the ball is thrown horizontally, the initial horizontal velocity remains constant throughout the motion. Therefore, the horizontal speed of the ball does not change.
Now, to determine the final horizontal velocity of the ball when it reaches the ground, we can use the horizontal distance it travels in 1.02 seconds. Since it was thrown horizontally, this distance would be equal to the initial horizontal velocity multiplied by the time.
Given that the initial horizontal velocity is 5 m/s and the time is 1.02 seconds, the equation would be:
Distance = Velocity * Time
Distance = 5 m/s * 1.02 s
Distance ≈ 5.1 m
So, the ball will be moving at a speed of approximately 5.1 m/s when it contacts the ground.
Problem 2:
We have a 70 kg pole-vaulter who converts the kinetic energy of running at ground level into potential energy to clear a crossbar at a height of 4 m. We want to find out the minimum velocity he must have had at takeoff to clear the bar, assuming his velocity at the top is zero.
To solve this problem, we can use the principle of conservation of energy.
1. At takeoff:
The total mechanical energy (kinetic energy + potential energy) of the pole-vaulter is conserved. At takeoff, the pole-vaulter would only have kinetic energy, which is given by:
KE = (1/2) * m * v^2
where KE is the kinetic energy, m is the mass, and v is the velocity.
2. At the top of the jump:
When the pole-vaulter reaches the top, the kinetic energy becomes zero, and there is only potential energy. The potential energy is given by:
PE = m * g * h
where PE is the potential energy, m is the mass, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height.
Since the total mechanical energy is conserved, we can equate the initial kinetic energy to the final potential energy. Therefore:
(1/2) * m * v^2 = m * g * h
Canceling out the mass, we get:
(1/2) * v^2 = g * h
Rearranging the equation:
v^2 = 2 * g * h
Taking the square root of both sides:
v = √(2 * g * h)
Substituting the given values, we get:
v = √(2 * 9.8 * 4)
v ≈ √78.4
v ≈ 8.85 m/s
So, the pole-vaulter must have a minimum velocity of approximately 8.85 m/s at takeoff in order to clear the 4 m high crossbar.
(1/2) m v^2 = m g h
v^2 = 2 gh
v = sqrt (2 g h)
= sqrt(2*9.81*4)
= 8.86 m/s