An athlete performing a long jump leaves the ground at a 27.0 degree angle and lands 7.80m away. what was the takeoff speed?

The initial velocity is 9.72 m/s.

Since an athlete performs a projectile motion;

R = u^2sin2θ/g

R = u^2sin2θ/g

Rg = u^2sin2θ

u^2 =Rg/sin2θ

u = √Rg/sin2θ

u = √7.80 × 9.8/sin(2)27.0

u = 9.72 m/s

The initial velocity is 9.72 m/s.

Learn more about projectiles: link

=9.72 m/s

Explanation:

From the Newton's laws of motion;

x=2(v²cos∅sin∅)/g

Using geometry we see that 2 cos∅sin∅ = sin 2∅

Therefore, x= (v²sin 2∅)g,  where v is the take off speed x the range and ∅ the launch angle.

Making v the subject of the formula we obtain the following equation.

v=√{xg /(sin 2∅)}

x=7.80

∅=27.0

v=√{7.8×9.8/sin(27×2)}

v=√94.485

v=9.72 m/s

e) none of the above answers are correct

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.