
Since initial velocity is zero hence , u = 0
=> d = 1/2 * a * t2

on solving we get
d = 86.436 metres
Note ; Here Gravitational Acceleration is take as , g = 9.8 m/s2
Answer:

Explanation:
Given


Required
Determine the mass of the ball
This question will be answered using Newton's second law of motion.
Which states that

Substitute values for Force and Acceleration

Make Mass the subject


--- (approximated)
Answer:
I=1,2•10³ kg•m/s
Explanation:
v¹=3.5m/s
vf=5m/s
v=5-3.5=1.5m/s
I=p
I=mv=850•1.5=1275 kg•m/s=1,2•10³ kg•m/s
Answer:
823.46 kgm/s
Explanation:
At 9 m above the water before he jumps, Henri LaMothe has a potential energy change, mgh which equals his kinetic energy 1/2mv² just as he reaches the surface of the water.
So, mgh = 1/2mv²
From here, his velocity just as he reaches the surface of the water is
v = √2gh
h = 9 m and g = 9.8 m/s²
v = √(2 × 9 × 9.8) m/s
v = √176.4 m/s
v₁ = 13.28 m/s
So his velocity just as he reaches the surface of the water is 13.28 m/s.
Now he dives into 32 cm = 0.32 m of water and stops so his final velocity v₂ = 0.
So, if we take the upward direction as positive, his initial momentum at the surface of the water is p₁ = -mv₁. His final momentum is p₂ = mv₂.
His momentum change or impulse, J = p₂ - p₁ = mv₂ - (-mv₁) = mv₂ + mv₁. Since m = Henri LaMothe's mass = 62 kg,
J = (62 × 0 + 62 × 13.28) kgm/s = 0 + 823.46 kgm/s = 823.46 kgm/s
So the magnitude of the impulse J of the water on him is 823.46 kgm/s
Answer:
100
Explanation:
by dividing 2000N and 1000kg.