-A person throws a ball upward into the air with an initial velocity of 15 m per second.
- How could you calculate how high the ball goes and how long the ball is in the air before it comes to the thrower's hand?
It travels upwards until the vertical velocity component becomes zero. An easier way to calculate it it to require that the initial kinetic energy become completely potential energy there.
(1/2) M Vo^2 = M g H
where H is the maximum height above the thrower's hand.
H = Vo^2/(2 g) = 11.5 m
The time in the air is twice the time it takes for the velocity upward to become zero. It spends an equal length of time coming back down. Therefore:
T = 2 * (Vo/g) = 3.06 s
thanx so much! im finally starting to get this!
To calculate how high the ball goes and how long it is in the air, we can use the kinematic equations of motion.
Step 1: Identify the known values:
- Initial velocity (u): 15 m/s (upward)
- Final velocity (v): 0 m/s (at the highest point)
- Acceleration (a): -9.8 m/s^2 (gravity, acting downwards)
Step 2: Calculate the time taken to reach the highest point:
Use the equation: v = u + at
0 = 15 - 9.8t
9.8t = 15
t = 15/9.8
t ≈ 1.53 seconds
Step 3: Calculate the height reached by the ball:
Use the equation: v^2 = u^2 + 2as
0^2 = 15^2 + 2(-9.8)s
0 = 225 - 19.6s
19.6s = 225
s = 225/19.6
s ≈ 11.48 meters
So, the ball reaches a height of approximately 11.48 meters and it takes approximately 1.53 seconds to reach the highest point.
Step 4: Calculate the time for the ball to fall back to the thrower’s hand:
Since the time taken to reach the highest point is the same as the time taken to fall back to the thrower's hand:
The total time in the air would be approximately 1.53 seconds (2 times the time calculated in Step 2).
Therefore, the ball is in the air for approximately 1.53 seconds before it comes back to the thrower's hand.
To calculate how high the ball goes and how long it is in the air before it comes to the thrower's hand, you can use the equations of motion. First, let's define the variables:
- Initial velocity (u) = 15 m/s (upward)
- Final velocity (v) = 0 m/s (at the top of the trajectory when it changes direction)
- Acceleration (a) = -9.8 m/s^2 (due to gravity, acting in the opposite direction of motion)
- Time taken (t) = unknown
- Displacement (s) = unknown
To find how high the ball goes, we can use the equation:
v^2 = u^2 + 2as
Since the final velocity is zero at the top of the trajectory, the equation becomes:
0^2 = u^2 + 2as
Simplifying the equation, we get:
u^2 = -2as
Substituting the given values:
15^2 = -2(-9.8)s
From this equation, you can solve for the displacement (s), which represents the height the ball reaches.
To find how long the ball is in the air, we can use the equation of motion:
v = u + at
Since the final velocity is zero when the ball comes back to the thrower's hand, the equation becomes:
0 = u + at
Simplifying the equation, we get:
u = -at
Substituting the given values:
15 = -9.8t
From this equation, you can solve for the time taken (t).
Note: The negative sign in the equations is due to the fact that the initial velocity is upward and the acceleration due to gravity is downward.