Answer:
He could jump 2.6 meters high.
Explanation:
Jumping a height of 1.3m requires a certain initial velocity v_0. It turns out that this scenario can be turned into an equivalent: if a person is dropped from a height of 1.3m in free fall, his velocity right before landing on the ground will be v_0. To answer this equivalent question, we use the kinematic equation:

With this result, we turn back to the original question on Earth: the person needs an initial velocity of 5 m/s to jump 1.3m high, on the Earth.
Now let's go to the other planet. It's smaller, half the radius, and its meadows are distinctly greener. Since its density is the same as one of the Earth, only its radius is half, we can argue that the gravitational acceleration g will be <em>half</em> of that of the Earth (you can verify this is true by writing down the Newton's formula for gravity, use volume of the sphere times density instead of the mass of the Earth, then see what happens to g when halving the radius). So, the question now becomes: from which height should the person be dropped in free fall so that his landing speed is 5 m/s ? Again, the kinematic equation comes in handy:

This results tells you, that on the planet X, which just half the radius of the Earth, a person will jump up to the height of 2.6 meters with same effort as on the Earth. This is exactly twice the height he jumps on Earth. It now all makes sense.
Below are the choices that can be found elsewhere:
A. (4.9 × 10-14 newtons) · tan(30°)
<span>B. (4.9 × 10-14 newtons) · sin(30°) </span>
<span>C. (4.9 × 10-14 newtons) · cos(30°) </span>
<span>D. (4.9 × 10-14 newtons) · arctan(30°) </span>
<span>E. (4.9 × 10-14 newtons) · arccos(30°)
</span>
<span>Force is proportional to the angle made by the velocity with respect to the magnetic field. It is maximum when velocity is perpendicular to the magnetic field and minimum when the velocity is parallel to the magnetic field. It is proportional to sin of the angle. In this problem it will be proportional to sin(30)</span>
Looks like you need to review through the lesson and take notes as it tells you in the lesson what each of these are.
Distance= speed (multiplied by) time
let the height of the person with marshmallow on her head be "h"
consider the motion of the marshmallow after it is dropped from bridge.
Y₀ = initial position of the marshmallow above the ground = 5.71 m
Y = final position of marshmallow on head of person = h
v₀ = initial velocity of the marshmallow = 0 m/s
a = acceleration due to gravity = - 9.8 m/s²
t = time of travel for marshmallow = 0.921 sec
Using the kinematics equation
Y = Y₀ + v₀ t + (0.5) a t²
inserting the values
h = 5.71 + 0 (0.921) + (0.5) (-9.8) (0.921)²
h = 5.71 - 4.16
h = 1.55 m