In the absence of air resistance, a ballplayer tosses a ball
straight up.
(a) By how much does the speed of the ball decrease
each second while it is ascending?
(b) By how much does its speed increase each second
while it is descending?
(c) How does the time of ascent compare with the time
of descent?
a. V = Vo-g*t
V = Vo-9.8m/s^2*1s
V = Vo-9.8m/s. s.
The speed decreases by 9.8 m each second.
b. V = Vo + g*t
V = Vo + 9.8m/s^2*1s
V = Vo + 9.8m/s.
The speed increases by 9.8 m each second.
c. V = Vo + g*t = 0 @ max. ht.
Vo + g*t = 0
Vo - 9.8t = 0
9.8t = Vo
Tr = Vo/9.8 = Rise time=Time of ascent.
V = Vo + g*t
V = 0 + 9.8*t
Tf = V/9.8
V = Vo during ascent.
Therefore, Tf = Vo/9.8 = Tr.
To answer these questions, we can use the principles of projectile motion and the equations of motion. Let's go step by step:
(a) By how much does the speed of the ball decrease each second while it is ascending?
During the ascent, the acceleration due to gravity acts against the ball's initial velocity. The value of acceleration is approximately -9.8 m/s² as it opposes the upward motion.
Using the equation of motion: v_f = v_i + at
Where:
v_f = final velocity
v_i = initial velocity
a = acceleration
t = time
Since the ball is thrown straight up, its initial velocity will be in the upwards direction and equal to the final velocity when it reaches its highest point (zero velocity).
So, the final velocity (v_f) can be taken as zero, and the acceleration (a) is -9.8 m/s².
0 = v_i - 9.8t
Solving for v_i, we get v_i = 9.8t.
Therefore, the speed of the ball decreases by 9.8 m/s each second while it is ascending.
(b) By how much does its speed increase each second while it is descending?
During the descent, the ball is under the influence of gravity, which accelerates it downwards. The acceleration is still approximately -9.8 m/s².
Using the same equation of motion: v_f = v_i + at
Here, the initial velocity (v_i) is zero since the ball reaches its highest point, where the velocity becomes zero before descending.
So, v_f = 0 and a = -9.8 m/s²
0 = 0 - 9.8t
Solving for t, we find t = 0.
Therefore, the speed does not increase during the descent as the initial velocity is zero.
(c) How does the time of ascent compare with the time of descent?
In projectile motion, the time taken to reach the highest point (time of ascent) is equal to the time taken to come back down (time of descent). This is true if air resistance is negligible.
Therefore, the time of ascent is equal to the time of descent.
To answer these questions, you can use a few basic principles of motion and equations of motion. Let's break it down step by step:
(a) By how much does the speed of the ball decrease each second while it is ascending?
The ball is subjected to the force of gravity while ascending, which causes its speed to decrease. The rate at which the speed decreases can be determined using the equation for velocity under constant acceleration:
v = u + at
Where:
v = final velocity
u = initial velocity
a = acceleration
t = time
In this case, the ball is tossed straight up, so its initial velocity, u, is positive. The acceleration, a, is the acceleration due to gravity, which is approximately -9.8 m/s^2 (negative because it acts in the opposite direction). The final velocity, v, is 0 at the peak of the ball's motion.
Substituting the values into the equation, we have:
0 = u - 9.8t
Solving for t, we get:
t = u / 9.8
So, the time it takes for the ball to reach its peak is given by u divided by 9.8.
(b) By how much does its speed increase each second while it is descending?
On the way down, the ball is still affected by the force of gravity but in the opposite direction. In this case, the acceleration, a, remains the same (-9.8 m/s^2), but the initial velocity, u, is 0 (as the ball reaches the peak and starts descending). The final velocity, v, is the velocity at any given time during the downward motion.
Using the same equation as before:
v = u + at
Substituting the values:
v = 0 - 9.8t
Simplifying, we find that the velocity of the ball while descending is given by:
v = -9.8t
Here, the negative sign indicates that the velocity is in the opposite direction to the initial upward motion.
(c) How does the time of ascent compare with the time of descent?
From our previous calculations, we found that the time to reach the peak, or the time of ascent, is given by t = u / 9.8. The time to descend can be found by rearranging the equation for velocity (v = -9.8t) to solve for t:
t = -v / 9.8
Since the velocity is negative during the descent, we can substitute -v for the velocity:
t = (-(-v)) / 9.8
t = v / 9.8
Comparing the expressions for ascent and descent times, we see that they are the same, except for a negative sign. Therefore, the time it takes for the ball to ascend is equal to the time it takes for the ball to descend.