Answer:
2.45 m/s
Explanation:
From the law of conservation of momentum,
The total momentum before the grab = Total momentum after the grab.
mu+m'u' = V(m+m')................... Equation 1
Where m = mass of Erica, m' = mass of Danny, u = initial velocity of Erica, u' = initial velocity of Danny, V = common velocity after garb
Make V the subject of the equation
V = (mu+m'u')/(m+m')............. Equation 2
Given: m = 37 kg, m' = 49 kg, u = 0 m/s(at the maximum height), u' = 4.3 m/s
Substitute into equation 2
V = (37×0+49×4.3)/(37+49)
V = 210.7/86
V = 2.45 m/s
Hence their speed just after the grab = 2.45 m/s
Answer: 2.45 m/s
Explanation:
To solve this, we use formula for law of conservation of momentum, in which, no momentum is lost in the system.
The initial momentum is equal to the final momentum
m(i)v(i) = m(f)v(f)
Erica's momentum is zero, while Danny's momentum is 49 kg * 4.3 m/s
Danny's momentum = 210.7 kgm/s
Now, using law of conservation of momentum, we have
210.7 kgm/s = (37 + 49) kg * v(f)
210.7 kgm/s = 86 kg * v(f)
v(f) = 210.7 kgm/s / 86 kg
v(f) = 2.45 m/s
Therefore, their speed just after he grabs her is 2.45 m/s
