A student throws a baseball vertically upward, and 2.8s later catches it at the same level. Neglecting air resistance , calculate the following
(a) the velocity at which the ball left the students hand
(b) the height to which the ball climbed above the students hand
(a) The baseball reached maximum height and zero velocity after t = 1.4 seconds, half the total interval. The initial velocity was Vo = g*t = 13.72 m/s
(b) The height reached is
H = (1/2) Vo t = Vo^2/(2g)
To solve this problem, we can use the equations of motion for an object in free fall. Since there is no air resistance, we can assume that the acceleration due to gravity is constant at -9.8 m/s².
(a) To find the velocity at which the ball left the student's hand, we can use the equation:
v = u + at,
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken.
Since the ball is thrown vertically upward, the final velocity (v) will be 0 m/s when the ball reaches its maximum height. The acceleration (a) is -9.8 m/s² (negative because it is acting in the opposite direction of the initial velocity). The time (t) is given as 2.8 s.
Substituting the values into the equation:
0 = u - 9.8 * 2.8
Simplifying the equation:
u = 9.8 * 2.8
u ≈ 27.44 m/s
Therefore, the velocity at which the ball left the student's hand is approximately 27.44 m/s.
(b) To find the height to which the ball climbed above the student's hand, we can use the equation:
s = ut + (1/2)at²,
where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time taken.
Since the ball reaches its maximum height at t = 2.8 s, we can substitute these values into the equation:
s = 27.44 * 2.8 + (1/2) * (-9.8) * (2.8)²
Simplifying the equation:
s = 27.44 * 2.8 - 4.9 * 2.8²
s = 76.832 - 38.528
s ≈ 38.304 m
Therefore, the height to which the ball climbed above the student's hand is approximately 38.304 meters.
To solve this problem, we'll use the equations of motion for an object under constant acceleration. In this case, the acceleration is due to gravity and is constant near the surface of the Earth.
Let's break down the problem using the given information:
(a) To find the velocity at which the ball left the student's hand, we need to use the equation:
v = u + at
where:
- v is the final velocity,
- u is the initial velocity,
- a is the acceleration, and
- t is the time interval.
Since the ball is thrown vertically upward, it is subject to an acceleration equal to the acceleration due to gravity (approximately 9.8 m/s^2), but in the opposite direction.
Given:
- The time interval t = 2.8 s.
Now, let's find the final velocity v:
v = u + at
0 = u - 9.8 * 2.8
Solving for u, we have:
u = 9.8 * 2.8
Therefore, the velocity at which the ball left the student's hand is approximately 27.4 m/s upward.
(b) To find the height to which the ball climbed above the student's hand, we can use the equation for displacement:
s = ut + 0.5at^2
where s is the displacement.
Considering that the ball reaches its highest point when its final velocity is zero, we can use this equation to find the height it reached.
Given:
- initial velocity u = 27.4 m/s (upward),
- time interval t = 2.8 s.
Substituting these values into the equation, we have:
s = ut + 0.5at^2
s = 27.4 * 2.8 + 0.5 * (-9.8) * (2.8)^2
Simplifying the equation:
s ≈ 38.6 m
Therefore, the ball climbed approximately 38.6 meters above the student's hand.