V=d*t
Since d=75km=75000m
t=1.5h=3600*1.5= 5400s
=>V=75000*5400
=405000000m/s=4.05*10^8m/s
Answer:
The final velocity of the second player is 6.1 m/s.
Explanation:
The final velocity of the second player can be calculated by conservation of linear momentum (p):
(1)
Where:
: is the mass of the first football player = 110 kg
: is the mass of the second football player = 90 kg
: is the initial velocity of the first football player = 5.0 m/s
: is the initial velocity of the second football player = 0 (he is at rest)
: is the final velocity of the first football player = 0 (he stops after the impact)
: is the final velocity of the second football player =?
By solving equation (1) for we have:
Therefore, the final velocity of the second player is 6.1 m/s.
I hope it helps you!
Explanation:
Given:
v₀ₓ = 15 m/s cos 20° = 14.10 m/s
aₓ = 0 m/s²
v₀ᵧ = 15 m/s sin 20° = 5.13 m/s
aᵧ = -9.8 m/s²
t = 1.5 s
Find: Δx and Δy
Δx = v₀ₓ t + ½ aₓ t²
Δx = (14.10 m/s) (1.5 s) + ½ (0 m/s²) (1.5 s)²
Δx = 21.1 m
Δy = v₀ᵧ t + ½ aᵧ t²
Δy = (5.13 m/s) (1.5 s) + ½ (-9.8 m/s²) (1.5 s)²
Δy = -3.33 m
