Given: Initial velocity

Final Velocity

To find : Height when jumpled 8.0 m/s upwards.

Solution: We already have values of initial velocity and final velocity.

We know, accereration due to gravity is given by .

It's negative because when jump it's in opposite direction.

Where h is the height when jumpled 8.0 m/s upwards.

Plugging values of

64= 225 -19.6h

Subtracting both sides by 225.

64-225= 225 -19.6h-225.

We get,

-161 = -19.6h

Dividing both sides by -19.6, we get

h= 8.2143

**Rounding to nearest tenth**, we get

**h= 8.2 meter.**

**His height is 8.2 meter when he is jumping 8.0 m/s upwards.**

Vf = Final velocity.

Vi = initial velocity

a = acceleration.

t = time

Vf = Vi + at

Vf = 0 + (2 m/s^2)(3s)

Vf = 6 m/s south

**Answer:**

**u = 3.35 m/s**

**Explanation:**

given,

mass , m = 0.455 kg

R = 0.675 m

Height of Loop = 1.021 m

the speed required at the top of loop be v

equating the force vertically

v² = 6.622

v = 2.57 m/s

Let the initial speed of ball be u

using conservation of energy

where,

0.7 u² = 7.85092

u² = 11.2156

**u = 3.35 m/s**

**the initial speed is 3.35 m/s**

**Answer:**

222.3 s

**Explanation:**

From Newton’s law, F=ma hence

Time,

Where is change in velocity

The change in velocity is equivalent to 0.106 m/s