You throw a ball vertically upward from the roof of a tall building. The ball leaves your hand at a point even with the roof railing with an upward speed of 15.0 m/s; the ball is then in free fall. On its way back down, it just misses the railing. At the location of the building, g = 9.80 m/s 2. Find (a) the position and velocity of the ball 1.00 s and 4.00 s after leaving your hand; (b) the velocity when the ball is 5.00 m above the railing; (c) the maximum height reached and the time at which it is reached; and (d) the acceleration of the ball when it is at its maximum height.
Answers
Explanation:
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PLZ MARK AS BRAINLIEST
Given:
Initial speed of the ball in upward direction = 12 m/s
g = 9.8 m/s²
To find:
(a) the position and velocity of the ball 1.00 s and 4.00 s after leaving your hand;
(b) the velocity when the ball is 5.00 m above the railing;
(c) the maximum height reached and the time at which it is reached; and
(d) the acceleration of the ball when it is at its maximum height.
Solution:
(a) The position and velocity of the ball 1.00 s and 4.00 s after leaving your hand;
Using the formula S = ut + 1/2 at², we get:
For t = 1 sec
S = 15*1 - 1/2*9.8*1²
S = 15 - 4.9
S = 10.1 m (over the railing)
For t = 4 sec
S = 15*4 - 1/2*9.8*4²
S = 60 - 78.4
S = -18.4 m (negative sign shows position of ball is below the railing)
(b) the velocity when the ball is 5.00 m above the railing;
Using the formula: v = u -√2gh
Where v is the final velocity of ball above the railing
u is the initial velocity
h is the height of the ball above the railing
Putting the values we get:
v = 15 - √2*9.8*5
v = 15 - √98
v = 5.1 m/s
(c) the maximum height reached and the time at which it is reached;
The kinetic energy of the ball goes in raising its height
Therefore 1/2*mv² = mgh
h = mv²/2mg
h = v²/2g = 15²/2*9.8 = 11.48 m (above the railing)
To find the time at which this height is attained:
v = u - at
Here v = 0
u = 15 m/s
a = g = 9.8 m/s²
Putting the values we get:
t = u/g = 15/9.8 = 1.531 sec
(d) the acceleration of the ball when it is at its maximum height.
The acceleration of the ball will just be the acceleration due to gravity acting on the ball which is 9.8 m/s²