A soccer ball is kicked at a speed of 25 m/s at 71 degrees. What is the velocity when the ball reaches a vertical height of 10 m?
-
This is one of those questions where you need to resolve the launch velocity of the ball into its vertical and horizontal components and look at each component separately.
Sketch the diagram.
The vertical component of velocity at launch = 25.sin(71°) = 23.64 m/s
We can now use one of the standards "suvat" kinematics equations to find the vertical component of its velocity at a height of 10 m.
v² = u² + 2as
v is the final velocity; u is the initial velocity; a is the acceleration; s is the distance
u = 23.64 m/s
s = 10 m
The acceleration is the acceleration due to gravity {"g"} I choose to take upwards as positive, so:
a = -9.81 m/s²
v² = 23.64² + (2 * -9.81 * 10)
v = 19.0 m/s
So at a height of 10m the vertical component of velocity is 19 m/s
The horizontal component of velocity at launch = 25.cos(71°) = 8.14 m/s
In this question, we are ignoring air resistance, so that will be the ball's horizontal component of velocity throughout the flight.
So at 10 m high, we have a ball with a vertical component of velocity of 19 m/s and a horizontal component of velocity of 8.14 m/s.
So we just need to find the resultant of those two components. Sketch the vectors and you see it's a job for Pythagoras.
v² = 19² + 8.14²
v = 20.67 m/s
We haven't quite finished yet. You are asked for a velocity, which has a magnitude {20.67 m/s} and a direction. From your sketch, you should see that the angle is the angle who's tangent is
(19 / 8.14) to the horizontal.
arc tan (2.33) = 66.8°
Velocity at 10 m = 20.67 m/s at 66.8°
Hope that helps
Sketch the diagram.
The vertical component of velocity at launch = 25.sin(71°) = 23.64 m/s
We can now use one of the standards "suvat" kinematics equations to find the vertical component of its velocity at a height of 10 m.
v² = u² + 2as
v is the final velocity; u is the initial velocity; a is the acceleration; s is the distance
u = 23.64 m/s
s = 10 m
The acceleration is the acceleration due to gravity {"g"} I choose to take upwards as positive, so:
a = -9.81 m/s²
v² = 23.64² + (2 * -9.81 * 10)
v = 19.0 m/s
So at a height of 10m the vertical component of velocity is 19 m/s
The horizontal component of velocity at launch = 25.cos(71°) = 8.14 m/s
In this question, we are ignoring air resistance, so that will be the ball's horizontal component of velocity throughout the flight.
So at 10 m high, we have a ball with a vertical component of velocity of 19 m/s and a horizontal component of velocity of 8.14 m/s.
So we just need to find the resultant of those two components. Sketch the vectors and you see it's a job for Pythagoras.
v² = 19² + 8.14²
v = 20.67 m/s
We haven't quite finished yet. You are asked for a velocity, which has a magnitude {20.67 m/s} and a direction. From your sketch, you should see that the angle is the angle who's tangent is
(19 / 8.14) to the horizontal.
arc tan (2.33) = 66.8°
Velocity at 10 m = 20.67 m/s at 66.8°
Hope that helps