A soccer player kicks a ball at an angle 45degree with an initial speed of 20m/s,assuming that the ball travels in a vertical plane. Calculate (a) The time at which the ball reaches the highest point. (b) The maximum height reached. (c) The horizontal range of the ball and (d) The time for which the ball is in the air.
a. Vo = 20m/s[45o].
Xo = 20*Cos45 = 14.14 m/s.
Yo = 20*sin45 = 14.14 m/s.
Y = Yo + g*t = 0.
14.14 - 9.8t = 0, t = 1.44 s.
b. Y^2 = Yo^2 + 2g*h = 0.
(14.14)^2 - 19.6h = 0, h = ?.
c. Range = Vo^2*sin(2A)/g. A = 45o.
d. T = Tr+Tf. Tf = Tr, T = Tr+Tr = 1.44 + 1.44 = 2.88 s.
d.
To find the answers to these questions, we can use the equations of motion for projectile motion. Here's how we can calculate each value:
(a) The time at which the ball reaches the highest point:
In projectile motion, the vertical component of the initial velocity is affected by gravity, while the horizontal component remains constant.
The time taken for an object to reach its highest point is equal to half of the total time of flight. The equation to find the time of flight is given by:
time_of_flight = 2 * (vertical_component_of_initial_velocity) / (acceleration_due_to_gravity)
Since the initial velocity is given as 20m/s and the angle is 45 degrees, we need to find the vertical component of the initial velocity.
vertical_component_of_initial_velocity = initial_velocity * sin(angle)
So, substitute the values into the equation:
vertical_component_of_initial_velocity = 20 * sin(45)
time_of_flight = 2 * (20 * sin(45)) / (9.8) = 2.04 seconds
Therefore, the time taken by the ball to reach the highest point is approximately 2.04 seconds.
(b) The maximum height reached:
In projectile motion, the maximum height reached by the object can be calculated using the equation:
maximum_height = (vertical_component_of_initial_velocity)^2 / (2 * acceleration_due_to_gravity)
Substitute the values:
maximum_height = (20 * sin(45))^2 / (2 * 9.8) = 20.41 meters
Therefore, the maximum height reached by the ball is approximately 20.41 meters.
(c) The horizontal range of the ball:
The horizontal range is the total horizontal distance covered by the ball.
horizontal_range = horizontal_component_of_initial_velocity * time_of_flight
Since the horizontal component of the initial velocity remains constant throughout the motion and is given by:
horizontal_component_of_initial_velocity = initial_velocity * cos(angle)
Substitute the values:
horizontal_component_of_initial_velocity = 20 * cos(45)
horizontal_range = (20 * cos(45)) * 2.04 = 40.82 meters
Therefore, the horizontal range of the ball is approximately 40.82 meters.
(d) The time for which the ball is in the air:
The total time of flight is the time during which the ball remains in the air.
time_of_flight = 2 * (vertical_component_of_initial_velocity) / (acceleration_due_to_gravity)
Substitute the values:
time_of_flight = 2 * (20 * sin(45)) / (9.8) = 2.04 seconds
Therefore, the time for which the ball is in the air is approximately 2.04 seconds.