Carlos is in an roof of a building of 20 meters tall, he throws a tennis ball vertically upwards at 15 m/s velocity after it reaches its highest height, the ball goes down until it reaches the ground (see the figure). Based on this information, answer the following.
a) Time taken to reach the highest height
b) The highest height in relation to building base
c) Total flight time since the ball is thrown until it reaches the ground
d) Its velocity when it crashes to the ground
a. Vf = Vo + g*Tr = 0.
Tr = -Vo/g = -15/-9.8 = 1.53 s. = Rise time.
b. h1 = Vo*Tr + 0.5g*Tr^2 = Ht. above the roof. g = -9.8 m/s^2.
h = h1 + 20.
c. h = 0.5g*Tf^2.
h = h1 + 20.
g = 9.8 m/s^2.
Tf = Fall time.
Tf = ?
Tr + Tf = Total flight time.
d. Vf^2 = Vo^2 + 2g*h.
Vo = 0.
g = 9.8 m/s^2.
h = h1 + 20.
To answer these questions, we can use the equations of motion relating to uniformly accelerated motion. Let's break down each question and find the answers step by step:
a) Time taken to reach the highest height:
To calculate the time taken to reach the highest height, we can use the equation for vertical motion:
v = u + at,
where:
v = final velocity (0 m/s at the highest point),
u = initial velocity (15 m/s),
a = acceleration (acceleration due to gravity, -9.8 m/s^2), and
t = time.
We rearrange the equation to solve for time (t):
t = (v - u) / a.
Plugging in the given values, we have:
t = (0 - 15) / -9.8.
Evaluating the expression, we find:
t ≈ 1.53 seconds.
Therefore, it takes approximately 1.53 seconds for the ball to reach its highest height.
b) The highest height in relation to the building base:
The highest height the ball reaches is equal to the displacement during the ascending phase of its motion. We can use the second equation of motion in vertical motion:
s = ut + (1/2)at^2,
where:
s = displacement,
u = initial velocity,
a = acceleration, and
t = time.
Since the ball is thrown upwards, the initial velocity is positive (15 m/s) and the acceleration is negative (-9.8 m/s^2) due to gravity. Thus, the equation becomes:
s = 15t + (1/2)(-9.8)t^2.
At the highest height, the final velocity is 0 m/s, so we can set v = 0 in the equation. Then we solve for t:
0 = 15t + (1/2)(-9.8)t^2.
This equation is quadratic, and we can solve it by factoring or using the quadratic formula. In this case, we will use the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / 2a,
where:
a = (1/2)(-9.8),
b = 15, and
c = 0.
Plugging in the values, we get:
t = (-15 ± √(15^2 - 4(1/2)(-9.8)(0))) / 2(1/2)(-9.8).
Simplifying further, we find:
t = (-15 ± √(225)) / -9.8.
Taking the positive root:
t = (-15 + √225) / -9.8.
Evaluating the expression, we find:
t ≈ -0.31 seconds or t ≈ 3.06 seconds.
Since time cannot be negative in this case, we disregard the negative solution. Therefore, the ball takes approximately 3.06 seconds to reach the ground.