Answer:
a) u = 45.21 m/s
b) v = 21.7 m/s
Explanation:
Given:
- Initial velocity : u
- Final velocity : v
- Initial height y(0) = 24 m
- Range = 100 m
Find:
- Find the initial and final velocities u and v?
Solution:
- The range of the half projectile motion is given as:
Range = u* sqrt ( 2*y(0) / g)
u = Range* sqrt ( g / 2*y(0))
- Plug in the values and solve for u:
u = 100*sqrt(9.81 / 2*24)
u = 100*0.452078533
u = 45.21 m/s
- Use third equation of motion to calculate the final velocity of the ball when it hits the ground in y - direction:
v^2 - u_y^2 = 2*g*y(0)
- Where vertical component of initial velocity is zero u_y = 0
v^2 = 2*g*y(0)
v = sqrt ( 2*g*y(0))
- plug values in:
v = sqrt ( 2*9.81*24)
v = 21.7 m/s
Answer:
a) Initial velocity = 0 m/s
b) Final velocity at ground = 21.7 m/s
Explanation:
We consider only vertical motion for the required information.
The ball travels a vertical distance equal to the height of the building. See the attached file diagram.
At point A the initial velocity of the ball will be zero because it just starts from there. So
Initial velocity =Vi= 0 m/s
During its flight the ball travels a vertical distance equal to the height od the building.
Distance =S=24m
Gravitational acceleration = g= 9.8 m/sec2
Velocity at point B = Vf=?
Using
[tex]2gS=Vf^{2}-Vi^{2}[/tex]
[tex]==> 2[/tex]×[tex]9.8[/tex]×[tex]24[/tex]=[tex]Vf^{2} -0^{2}[/tex]
==> [tex]470.4=Vf^{2}[/tex]
==> Vf =[tex]\sqrt{470.4}[/tex]
==> Vf= 21.7 m/sec
So the ball will hit the ground with 21.7 m/s velocity
