a ball is thrown upward from a bridge with a speed of 48 ft/s. It misses the bridge on way down and lands in water 160 ft. below. Find how long the ball rises, how high it goes, how long it is in the air, and its velocity when it strikes the water?

h(t) = 160 + 48t - 16t^2

v(t) = 48 - 32t
See what you can do with that.

To find the answers to these questions, we can use the laws of motion and the equations of projectile motion. Let's break down each question and find the solutions step by step.

1. How long does the ball rise?
To find the time it takes for the ball to reach its maximum height, we can use the equation for the time of flight of a projectile in freefall. The initial upward velocity is 48 ft/s, and the acceleration due to gravity is -32 ft/s² (assuming upward as positive). We can use the equation:

t = v / g

where t is the time, v is the velocity, and g is the acceleration due to gravity. Plugging in the values in this case, we get:

t = 48 ft/s / 32 ft/s²
t = 1.5 seconds

Therefore, the ball takes 1.5 seconds to reach its maximum height.

2. How high does the ball go?
To find the maximum height reached by the ball, we can use the equation for displacement in vertical projectile motion. Since the ball goes up and then down, the total displacement would be twice the displacement from the launch point to the maximum height. We can use the equation:

h = (v^2) / (2 * g)

where h is the height, v is the velocity, and g is the acceleration due to gravity. Plugging in the values in this case, we get:

h = (48 ft/s)^2 / (2 * 32 ft/s²)
h = 36 ft

Therefore, the ball reaches a height of 36 feet.

3. How long is the ball in the air?
To find the total time the ball is in the air, we need to double the time it takes to reach its maximum height. Since we already found that the ball spends 1.5 seconds to reach the maximum height, we multiply that by 2:

Total time = 1.5 seconds * 2
Total time = 3 seconds

Therefore, the ball is in the air for a total of 3 seconds.

4. What is the velocity of the ball when it strikes the water?
To find the velocity of the ball when it strikes the water, we can use the equation for final velocity in vertical projectile motion. The final velocity can be found by adding the initial velocity to the product of the acceleration due to gravity and the time. We can use the equation:

vf = vi + (g * t)

where vf is the final velocity, vi is the initial velocity, g is the acceleration due to gravity, and t is the time. Plugging in the values in this case, we get:

vf = 48 ft/s + (-32 ft/s² * 3 s)
vf = -96 ft/s

Therefore, the velocity of the ball when it strikes the water is -96 ft/s (negative sign indicates the downward direction).

1. V = Vo + g*Tr = 0.,

48 + (-32)Tr = 0,
Tr36 + = 1.5 s. = Rise time.

2. V^2 = Vo^2 + 2g*h = 0.
48^2 + (-64)h = 0,
h = 36 ft.

3. h = 0.5g*Tf^2 = 36 + 160.
16*Tf^2 = 196,
Tf = 3.5 s. = Fall time.
Tr + Tf = 1.5 + 3.5 = 5.0 s. = Time in air.

4. V = Vo + g*Tf = 0 + 32*3.5 =