A ball is thrown vertically up and from the ground with a velocity of 20ms. Calculate:
a-The maximum length reached.
b-The time to reach the maximum height.
c-The time to reach the ground again after the ball is thrown up.
d-The velocity reached halfway to the maximum height.
Assure g=10ms.
g does NOT equal 10 ms. Is that supposed to be milliseconds?
20 ms is also not a velocity
Perhaps they want you to USE g = 10 m/s^2.
Not only are you showing no work, you are not even copying the problem correctly.
physics
To solve this problem, we can use kinematic equations of motion. Let's break down each part of the question step by step:
a) The maximum height reached by the ball can be found using the formula for vertical motion. In this case, the initial velocity is 20 m/s, and the acceleration due to gravity is -10 m/s² (negative because it is acting in the opposite direction). We can use the equation:
v² = u² + 2as
Here, v is the final velocity (which is 0 m/s at the maximum height), u is the initial velocity (20 m/s), a is the acceleration (-10 m/s²), and s is the maximum height we need to find.
Rearranging the equation, we have:
0² = 20² + 2(-10)s
0 = 400 - 20s
20s = 400
s = 400/20
s = 20 meters
Therefore, the maximum height reached by the ball is 20 meters.
b) The time to reach the maximum height can be found using the equation:
v = u + at
Here, v is the final velocity (0 m/s), u is the initial velocity (20 m/s), a is the acceleration due to gravity (-10 m/s²), and t is the time we need to find.
Rearranging the equation, we have:
0 = 20 - 10t
10t = 20
t = 20/10
t = 2 seconds
Therefore, the time taken to reach the maximum height is 2 seconds.
c) The time to reach the ground again after the ball is thrown up can be found using the equation:
s = ut + (1/2)at²
Since the ball's initial velocity is upwards, the displacement s is 20 meters (the maximum height), and acceleration a is -10 m/s², we can rearrange the equation and solve for t:
20 = 20(2) + (1/2)(-10)t²
20 = 40 - 5t²
5t² = 20-40
5t² = -20
t² = -20/5
t² = -4
Since time cannot be negative, the time to reach the ground again does not exist. This means that the ball will never reach the ground again after being thrown up.
d) The velocity reached halfway to the maximum height can be found using the equation:
v = u + at
Here, v is the final velocity that we need to find, u is the initial velocity (20 m/s), a is the acceleration due to gravity (-10 m/s²), and t is the time taken to reach halfway.
To find the time taken to reach halfway, we divide the time to reach the maximum height (2 seconds) by 2:
t_half = 2/2
t_half = 1 second
Substituting this value into the equation, we can solve for v:
v = 20 + (-10)(1)
v = 20 - 10
v = 10 m/s
Therefore, the velocity reached halfway to the maximum height is 10 m/s.