A shot put is thrown upward with a velocity of 35 ft./sec/ at a height of 4 ft. and an angle of 40 degrees. How long will it take for the shot put to be a horizontal distance of 40 ft. from the person throwing it? Acceleration due to gravity is 32 ft./s^2. Round your answer to the nearest hundredth.
Vo = 35Ft./s[40o]
Xo = 35*cos40 = 26.8 m/s.
d = Xo*T = 40
26.8*T = 40
T =
Is the answer 1.49?
Yes.
To solve this problem, we can break it down into two components: the horizontal motion and the vertical motion of the shot put.
First, let's focus on the horizontal motion. We need to find the time it takes for the shot put to travel a horizontal distance of 40 ft. Since there is no acceleration in the horizontal direction, we can use the formula:
distance = velocity * time
In this case, the distance is 40 ft. and the velocity is the horizontal component of the initial velocity. To find the horizontal component, we can use the formula:
horizontal velocity = initial velocity * cos(angle)
Plugging in the values, we have:
horizontal velocity = 35 ft./sec * cos(40 degrees)
Now we can solve for time:
40 ft. = (35 ft./sec * cos(40 degrees)) * time
Dividing both sides by (35 ft./sec * cos(40 degrees)), we get:
time = 40 ft. / (35 ft./sec * cos(40 degrees))
Now, let's focus on the vertical motion. We need to find the time it takes for the shot put to reach a height of 4 ft. Since the initial velocity is given in terms of horizontal and vertical components, we need to find the time it takes for the shot put to reach its maximum height, and then double that time to find the total time it takes to reach a height of 4 ft.
The formula for finding the time it takes for an object to reach its maximum height in vertical motion is:
time to max height = vertical velocity / acceleration due to gravity
In this case, the vertical velocity is the vertical component of the initial velocity. To find the vertical component, we can use the formula:
vertical velocity = initial velocity * sin(angle)
Plugging in the values, we have:
vertical velocity = 35 ft./sec * sin(40 degrees)
Now we can solve for the time to reach the maximum height:
time to max height = (35 ft./sec * sin(40 degrees)) / 32 ft./s^2
Since the shot put needs to reach a height of 4 ft., we need to double the time to max height:
total time = 2 * time to max height
Finally, we can substitute the values we calculated into the equation for the horizontal motion, to find the final answer for how long it will take for the shot put to be a horizontal distance of 40 ft. from the person throwing it:
time = 40 ft. / (35 ft./sec * cos(40 degrees)) + 2 * [(35 ft./sec * sin(40 degrees)) / 32 ft./s^2]
Using a calculator, we can plug in the values and calculate the answer.