a projectile is fired from the top of a building with an initial velocity straight upward at 40m/sec. the top of the building is 50 meters above the ground?
a) the maximun height of the projectile above the building
b) the time for the projectile to reach the top
c) the velocity of the projectile as hits the ground
d) the total time the projectile is in the air.
A. Max height = 81.549m ( above the building)
B. Time to reach max height = 4 seconds
C. Velocity = 50.80 meter per second
D. total time = 9.26 seconds
To solve these questions, we can use the equations of motion for projectile motion. Let's break down each part:
a) The maximum height of the projectile above the building:
To find the maximum height reached by the projectile, we need to determine the time it takes for the projectile to reach its peak. We can use the equation for vertical displacement:
y = y0 + v0y * t + (1/2) * a * t^2
Where:
y = final vertical displacement (maximum height above the building)
y0 = initial vertical displacement (starting height above the ground)
v0y = initial vertical velocity
a = acceleration due to gravity (approximately -9.8 m/s^2)
t = time taken to reach the maximum height
In this case, the initial vertical velocity (v0y) is 40 m/s (upward), the initial height (y0) is 50 meters (above the ground), and the vertical acceleration (a) is -9.8 m/s^2. Since the final displacement y is the maximum height, we can set it as the unknown. We can solve for t and substitute the values to find the maximum height.
b) The time for the projectile to reach the top:
The time taken to reach the top can be found using the equation for final velocity:
v = v0 + a * t
Where:
v = final vertical velocity
v0 = initial vertical velocity
a = acceleration due to gravity (approximately -9.8 m/s^2)
t = time taken to reach the top
In this case, the initial vertical velocity (v0) is 40 m/s (upward), and the vertical acceleration (a) is -9.8 m/s^2. The final velocity (v) at the top is 0 because the projectile comes to a stop before falling downward. We can solve for t using these values.
c) The velocity of the projectile as it hits the ground:
The velocity of the projectile as it hits the ground is the same as its initial velocity but in the opposite direction. So, the velocity would be -40 m/s (downward) in this case.
d) The total time the projectile is in the air:
To find the total time of flight, we can use the equation for vertical displacement:
y = y0 + v0y * t + (1/2) * a * t^2
Where:
y = final vertical displacement (0 when the object returns to the ground)
y0 = initial vertical displacement (starting height above the ground)
v0y = initial vertical velocity
a = acceleration due to gravity (approximately -9.8 m/s^2)
t = time taken for the projectile to return to the ground
In this case, the initial vertical velocity (v0y) is 40 m/s (upward), the initial height (y0) is 50 meters (above the ground), and the vertical acceleration (a) is -9.8 m/s^2. We can solve for t to find the total time of flight.
By applying these equations and substituting the given values, we can determine the answers to each part of the question.