a ball is thrown up and reaches a maximum height of 5 m.
a) what is the initial velocity of the ball?
b) how long does it take for the ball to reach its maximum height?
c) what is the final velocity of the ball? if i returns to where it was thrown from?
y = y0 + v0*t - 1/2*g*t^2
where y0 is the initial height = 0, v0 is the initial speed, and g is the acceleration due to gravity, t is time, and y is the vertical position.
The ball reaches it's maximum height when dy/dt = 0 = v0 - g*t
and y = 5, so
5 = v0*t - 1/2*g*t^2
Use algebra to find the initial velocity and time.
The final velocity of the ball is given by
1. v = v0 - g*t, occurs when y = 0, or when
2. 0 = v0*t - 1/2*g*t^2
solve 2nd equation for t, and plug into 1st equation to solve for the final speed v at this time.
a sandbag is dropped from an unknown height. h, from the ground. if after 5 sec. it is located one half the height from which is thrown , at what height was it dropped?
b. what is its speed after 5 sec.?
c. what is its speed when it strikes the ground?
To find the answers to these questions, we need to use the equations of motion for an object in free fall. Let's break it down step by step:
a) To find the initial velocity of the ball, we can use the equation for the maximum height reached during vertical motion:
Vf^2 = Vi^2 + 2gh
Given:
Vf = 0 (at the maximum height)
h = 5 m (maximum height above the starting point)
Rearranging the equation, we have:
Vi^2 = -2gh
Substituting the values, we get:
Vi^2 = -2 * 9.8 m/s^2 * 5 m
Simplifying, we find:
Vi^2 = -98 m^2/s^2
Taking the square root of both sides, we get:
Vi = ± √(-98 m^2/s^2)
Since velocity cannot be negative in this context, we ignore the negative sign:
Vi ≈ 9.9 m/s (rounded to one decimal place)
Therefore, the initial velocity of the ball is approximately 9.9 m/s.
b) To find the time it takes for the ball to reach its maximum height, we can use the equation for vertical motion:
Vf = Vi + gt
Given:
Vf = 0 (at the maximum height)
Vi = 9.9 m/s (initial velocity)
g = 9.8 m/s^2 (acceleration due to gravity)
Substituting the values, we have:
0 = 9.9 m/s + 9.8 m/s^2 * t
Rearranging the equation and solving for t, we find:
t = -9.9 m/s / 9.8 m/s^2
t ≈ -1 second
Since time cannot be negative in this context, we ignore the negative sign:
t ≈ 1 second
Therefore, it takes approximately 1 second for the ball to reach its maximum height.
c) To find the final velocity of the ball when it returns to its starting point, we can use the equation for vertical motion:
Vf = Vi + gt
Given:
Vf = 0 (at the starting point)
Vi = 9.9 m/s (initial velocity)
g = 9.8 m/s^2 (acceleration due to gravity)
Substituting the values, we have:
0 = 9.9 m/s + 9.8 m/s^2 * t
Rearranging the equation and solving for t, we find:
t = -9.9 m/s / 9.8 m/s^2
t ≈ -1 second
Since time cannot be negative in this context, we ignore the negative sign:
t ≈ 1 second
Therefore, it takes approximately 1 second for the ball to return to its starting point.
Since the final velocity at the starting point is zero, the final velocity of the ball is 0 m/s.