Answer:
a)
the quarterback will be moving back at speed of 0.080625 m/s
b)
the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm
Explanation:
Given the data in the question;
a)
How fast will he be moving backward just after releasing the ball?
using conservation of momentum;
m₁v₁ = m₂v₂
v₂ = m₁v₁ / m₂
where m₁ is initial mass ( 0.43 kg )
m₂ is the final mass ( 80 kg )
v₁ is the initial velocity ( 15 m/s )
v₂ is the final velocity
so we substitute
v₂ = ( 0.43 × 15 ) / 80
v₂ = 6.45 / 80
v₂ = 0.080625 m/s
Therefore, the quarterback will be moving back at speed of 0.080625 m/s
b) Suppose that the quarterback takes 0.30 to return to the ground after throwing the ball. How far d will he move horizontally, assuming his speed is constant?
we make use of the relation between time, distance and speed;
s = d/t
d = st
where s is the speed ( 0.080625 m/s )
t is time ( 0.30 s )
so we substitute
d = 0.080625 × 0.30
d = 0.0241875 m or 2.41875 cm
Therefore, the distance moved horizontally by the quarterback is 0.0241875 m or 2.41875 cm
