Question

You are standing 24.1 meters away from Brown Hall. After your physics exam you want to kick a ball at the building. You kick the ball with an initial velocity of 26.7 m/s and at an angle of 33 degrees above the horizontal. Give two decimal places for your answers. What is the x-component of the initial velocity, ∼m/s What is the y-component of the initial velocity, X m/s How much time does it take for the ball to reach the building? - seconds How high up the wall, does the ball hit the building? x meters

Answer 1

I apologize, it looks like my previous response was cut off. Here are the full answers to the questions:

The x-component of the initial velocity is given by:

Vx = V0 cosθ

where V0 is the initial velocity and θ is the angle above the horizontal. Substituting the given values, we get:

Vx = 26.7 cos(33°) = 22.35 m/s (to two decimal places)

Therefore, the x-component of the initial velocity is approximately 22.35 m/s.

The y-component of the initial velocity is given by:

Vy = V0 sinθ

Substituting the given values, we get:

Vy = 26.7 sin(33°) = 14.13 m/s (to two decimal places)

Therefore, the y-component of the initial velocity is approximately 14.13 m/s.

To find the time taken for the ball to reach the building, we can use the equation for the time of flight of a projectile:

t = 2Vy / g

where g is the acceleration due to gravity. Substituting the given values, we get:

t = 2(14.13) / 9.8 = 2.88 seconds (to two decimal places)

Therefore, it takes approximately 2.88 seconds for the ball to reach the building.

Tofind the height at which the ball hits the building, we can use the equation:

y = h + Vy t - 0.5 g t^2

where h is the initial height of the ball (which we can assume is zero), and y is the vertical distance traveled by the ball. Substituting the given values, we get:

y = 0 + 14.13(2.88) - 0.5(9.8)(2.88)^2 = 18.05 meters (to two decimal places)

Therefore, the ball hits the building at a height of approximately 18.05 meters above the ground.

Step-by-step explanation: