Answer:
19.3m/s
Explanation:
Use third equation of motion

where v is the velocity at halfway, u is the initial velocity, g is gravity (9.81m/s^2) and h is the height at which you'd want to find the velocity
insert values to get answer
![v^2-0^2=2(9.81m/s^2)(38/2)\\v^2=9.81m/s^2 *38\\v^2=372.78\\v=\sqrt[]{372.78} \\v=19.3m/s](https://tex.z-dn.net/?f=v%5E2-0%5E2%3D2%289.81m%2Fs%5E2%29%2838%2F2%29%5C%5Cv%5E2%3D9.81m%2Fs%5E2%20%2A38%5C%5Cv%5E2%3D372.78%5C%5Cv%3D%5Csqrt%5B%5D%7B372.78%7D%20%5C%5Cv%3D19.3m%2Fs)
Answer:

Explanation:
As we know that the formula of range is given as

now we know that
maximum value of the range of the projectile is given as

now we need to find such angles for which the range is half the maximum value
so we will have




We can use the equation for kinetic energy, K=1/2mv².
Your given variables are already in the correct units, so we can just plug in the variables and solve for v.
K = 1/2mv²
16 = 1/2(2)v²
16 = (1)v²
√16 = v
v = 4 m/s
Therefore, the velocity of a 2 kg mass with 16 J of kinetic energy is 4 m/s.
Hope this is helpful!
Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).