a ball thrown vertically upward lands on the thrower's hand 24.5s. find (a) the time taken by the ball to reach its maximum height,(b) the maximum height attained by the ball, and (c) initial velocity of the ball
goes up for 24.5/2 seconds
(by the way, no way! )
v = Vi - 9.81 t
v = 0 at top
so
Vi = -9.81(12.25)
h = Vi t - 4.9 t^2
h = Vi (12.5) - 4.9(12.5)^2
i dont no because i want a sulotion plz
To find the time taken by the ball to reach its maximum height, you can use the fact that the time it takes for an object to reach its peak height is equal to half of the total time of flight.
(a) The time taken by the ball to reach its maximum height is equal to half of the total time of flight, which is 24.5s / 2 = 12.25s.
To find the maximum height attained by the ball, you can use the kinematic equation for vertical motion:
h = (v^2 - u^2) / (2g)
where h is the height, v is the final velocity, u is the initial velocity, and g is the acceleration due to gravity.
(b) At the maximum height, the final velocity v is zero. The acceleration due to gravity, g, is approximately 9.8 m/s^2.
Using the equation with the given values:
0 = (0 - u^2) / (2 * (-9.8))
Simplifying the equation, we have:
0 = -u^2 / (-19.6)
Multiplying both sides by -19.6, we get:
0 = u^2
Therefore, the initial velocity u is 0 m/s.
(c) The initial velocity of the ball is 0 m/s.
To find the time taken by the ball to reach its maximum height, we need to know that when an object is thrown vertically upward, its velocity decreases until it reaches its highest point where the velocity becomes zero. After that, the object starts to fall back under the force of gravity.
(a) The time taken by the ball to reach its maximum height is half of the total time taken. In this case, the total time taken is 24.5 seconds. Therefore, the time taken to reach the maximum height is 24.5 / 2 = 12.25 seconds.
(b) To find the maximum height attained by the ball, we can use the equation for vertical motion:
h = u*t + (1/2)*g*t^2
Where:
h = height
u = initial velocity
t = time
g = acceleration due to gravity (-9.8 m/s^2)
At the maximum height, the final velocity is zero, so we can use this information to calculate the height.
0 = u - g*t
Solving for u, the initial velocity:
u = g*t
Substituting this value in the equation for height, we get:
h = (1/2)*g*t^2
Plugging in the values:
h = (1/2)*(-9.8)*(12.25^2)
Simplifying, we find:
h = 760.56 meters
So, the maximum height attained by the ball is 760.56 meters.
(c) Since the initial velocity is the same as the final velocity at the maximum height, we can use the equation we found earlier:
u = g*t
Substituting the values:
u = -9.8 * 12.25
Simplifying, we find:
u = -119.75 m/s (rounded to two decimal places)
So, the initial velocity of the ball is approximately -119.75 m/s.