A football is kicked from the ground with a speed of 23m/s at an angle of 42 degrees above the horizontal
a) how long is the football in the air
b) how far does it travel
First, break down the velocity into its' horizontal and vertical components:
Vx=23m/s*Cos42=17.09m/s
Vy=23m/s*Sin42=15.39m/s
Use the following kinematic equation:
Vf=Vi+at
Where
Vi=15.39m/s
a=g=-9.8m/s
t=???
and
Vf=0m/s
***When the football reaches it's peak, the velocity will be 0m/s
Solve for t using Vy for Vi:
0=15.39m/s+(-9.8m/s^2)t
-15.39m/s/-9.8m/s^2=t
t=1.57s
It takes the same amount of time for the ball to reach the top, as it takes to come back down. So, the total time is
2*t=2*(1.57s)=3.14s=3.1s
Joel and I agree.
I don't think that the author of the question want's to know the distance that the ball travels in the y-direction, but the author wants to know the distance in the horizontal direction. To solve for that, use the following equation:
d=Vx*t
d=17.09m/s*3.14s
d=53.67m=54m
A) How long, i suppose that's the time
Time, T = 2uSin42�0†2/g
T = 2 x 23 x 0.6691/10
T = 31.37/10
T = 3secs.
B) I suppose that's the height too.
Maximum height, H = u�0…5Sin�0…542�0†2/2g
H = 23�0…5 x (0.6691)�0…5/2 x 10
H = 529 x 0.4477/20
H = 236.8333/20
H = 11.84m
A shot-putter launches a 7.5kg shot with a speed of 12m/s at an angle of 37 degrees above the horizontal. if, upon release, the shot was 2.0m above the ground, determine (a) the horizontal range of the shot. (b) the speed of the projectile through the air just before it hits the ground.
To find the answers to both parts of this question, we can break down the motion of the football into its horizontal and vertical components.
First, let's determine the initial vertical velocity (Vy) and horizontal velocity (Vx) of the football using trigonometry.
Given:
Initial speed (V) = 23 m/s
Launch angle (θ) = 42 degrees
Vertical velocity (Vy) = V * sin(θ)
= 23 m/s * sin(42°)
≈ 14.68 m/s
Horizontal velocity (Vx) = V * cos(θ)
= 23 m/s * cos(42°)
≈ 17.42 m/s
a) How long is the football in the air?
We can use the vertical motion equation to find the time of flight (t) for the football. The equation is:
Δy = Vy * t + (1/2) * g * t^2
Where:
Δy = Vertical displacement (change in height)
Vy = Initial vertical velocity
g = Acceleration due to gravity (approximately 9.8 m/s^2)
t = Time
Since the football is launched and lands at the same height, Δy = 0. Therefore, the equation becomes:
0 = Vy * t + (1/2) * g * t^2
Substituting the known values:
0 = 14.68 m/s * t + (1/2) * 9.8 m/s^2 * t^2
Simplifying and rearranging the equation:
4.9 t^2 + 14.68 t = 0
Factoring out t:
t (4.9 t + 14.68) = 0
Setting each factor equal to zero gives us two solutions:
t = 0 (This isn't meaningful in this context)
4.9 t + 14.68 = 0
Solving for t:
4.9 t = -14.68
t = -14.68 / 4.9
t ≈ -3 seconds (This negative time does not have a physical meaning in this context)
Since time cannot be negative, we discard this solution. Therefore, the football is in the air for approximately 3 seconds.
b) How far does it travel?
We can use the horizontal motion equation to find the horizontal displacement (range) covered by the football. The equation is:
Δx = Vx * t
Where:
Δx = Horizontal displacement (range)
Vx = Initial horizontal velocity
t = Time of flight
Substituting the known values:
Δx = 17.42 m/s * 3 s
Calculating:
Δx ≈ 52.26 meters
Therefore, the football travels approximately 52.26 meters.