A skydiver is using wind to land on a target that is 120 m away horizontally. The skydiver starts from a height of 70 m and is falling vertically at a constant velocity of 7.0 m/s downward with their parachute open (terminal velocity). A horizontal gust of wind helps push them towards the target. What must be their total speed if they want to just hit their target

13.9 m/s.

Explanation:

Since the vertical velocity of the skydiver is constant at v = 7.0 m/s, we find the time, t it takes him to drop from a height of h = 70 m.

So, distance = velocity time

h = vt

t = h/v = 70 m/7 m/s = 10 s

This is also the time it takes him to move horizontally a distance of d = 120 m to the target.

So, his horizontal velocity is v' = distance/time = d/t = 120m/10 s = 12 m/s.

Since both vertical and horizontal velocities are perpendicular, we add them vectorially to obtain the skydivers total speed, V.

So, V = √(v² + v'²)

= √((7.0 m/s)² + (12.0 m/s)'²)

= √(49 m²/s² + 144 m²/s²)

= √(193 m²/s²)

= 13.9 m/s.

The direction of this velocity is Ф = tan⁻¹(v/v')

= tan⁻¹(7 m/s/12 m/s)

= tan⁻¹(0.5833)

= 30.3°

where's the following? you gotta add it to know what the choices are

explanation:

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s