Answer: a) To determine how long the ball is in the air, we can use the equation for the vertical motion of an object under constant acceleration due to gravity:
h(t) = h0 + v0t - (1/2)gt^2
where h(t) is the height of the ball at time t, h0 is the initial height (which we can assume to be 0 since the ball is thrown into the air), v0 is the initial velocity (20 ft/s), g is the acceleration due to gravity (32 ft/s^2), and t is the time in seconds.
Since the ball reaches its maximum height when its vertical velocity becomes 0, we can set v0t - (1/2)gt^2 = 0 and solve for t:
v0t - (1/2)gt^2 = 0
20t - 16t^2 = 0
4t(5 - 4t) = 0
Solving for t, we get t = 0 (which is the time when the ball is thrown) or t = 5/4 seconds. However, the time when the ball is thrown is not the time it is in the air, so we discard t = 0. Therefore, the ball is in the air for 5/4 seconds.
b) To determine how high the ball goes, we can use the equation for the maximum height reached by an object under constant acceleration due to gravity:
h_max = h0 + (v0^2)/(2g)
Plugging in the values for h0, v0, and g, we get:
h_max = 0 + (20^2)/(2 * 32)
h_max = 200/64
h_max = 3.125 ft
Therefore, the ball reaches a maximum height of 3.125 ft.
c) To determine the time when the ball is at a height of 3 ft, we can set h(t) = 3 and solve for t:
h(t) = 0 + 20t - 16t^2 = 3
16t^2 - 20t + 3 = 0
We can solve this quadratic equation for t using the quadratic formula:
t = (-b ± √(b^2 - 4ac)) / (2a)
where a = 16, b = -20, and c = 3. Plugging in these values, we get:
t = (-(-20) ± √((-20)^2 - 4 * 16 * 3)) / (2 * 16)
t = (20 ± √(400 - 192)) / 32
t = (20 ± √208) / 32
We can simplify √208 as approximately 14.387, so we get two possible values for t:
t ≈ (20 + 14.387) / 32 ≈ 1.388 seconds
t ≈ (20 - 14.387) / 32 ≈ 0.183 seconds
Therefore, the ball is at a height of 3 ft at approximately 1.388 seconds or 0.183 seconds.
To determine if you will hit the object before coming to a stop, we can calculate the stopping distance of the car and compare it to the range of your headlights. The stopping distance can be calculated using the equation:
d = (v^2) / (2a)
where d is the stopping distance, v is the initial velocity (100 km/h, which can be converted to m/s), and a is the deceleration due to braking (-7 m/s^2).
Converting the initial velocity to m/s:
v = 100 km/h * (1000 m / 1 km) * (1 h / 3600 s) ≈ 27.78 m/s
Plugging in the values for v and a into the stopping distance equation:
d = (27.78^2) / (2 * -7)
d ≈ 111.12 m
Therefore, the stopping distance of the car is approximately 111.12 meters.
Since the range of your headlights is 30 meters, and the stopping distance of the car is greater than 30 meters, the car will hit the object before coming to a stop.
b) The time it takes to stop can be calculated using the equation:
t = v / a
where t is the time to stop, v is the initial velocity (27.78 m/s), and a is the deceleration due to braking (-7 m/s^2).
Plugging in the values for v and a into the time to stop equation:
t = 27.78 / -7
t ≈ -3.97 s
The negative sign in the time indicates that the car is decelerating, or slowing down. However, time cannot be negative in this context, so we can take the absolute value to get the positive time:
|t| ≈ 3.97 s
Therefore, it will take approximately 3.97 seconds for the car to come to a complete stop.