Question

An object is thrown upwards with a speed of 14 m/s. How long does it take to reach a height of 5.0 m above the projection point while descending?

Answer 1

Answer:

2.43 s

Explanation:

Using newton's equation of motion.

T = (v-u)/g

Where T = time taken for the object to return to the point of projection , u = initial velocity, v = final velocity, g = acceleration due to gravity.

Given: v =-14 m/s, u = 14 m/s, g = -9.8 m/s²

T = (-14-14)/-9.81

T = 2.85 s

Note: We look for the object's speed at 5.0 m.

using

v² = u²+2gs.................................... Equation 1

Where v = final velocity, u = initial velocity, g = acceleration due to gravity, s = distance.

Given:  u = 14 m/s, g = -9.81 m/s², s = 5.0 m

Substitute into equation 1

v² = 14²+(-9.81×5×2)

v² = 196-98.1

v = √97.9

v = 9.89

We look for the time taken for the velocity to decrease from 14 m/s to 9.89 m/s.

using

v = u+gt

t =(v-u)/g........................... Equation 2

Where t = time taken for the object to decrease it velocity from 14 m/s to 9.89 m/s

Given: v = 9.89 m/s, u =14 m/s g = -9.81 m/s²

t = (14-9.89)/-9.81

t = -4.11/-9.81

t = 0.42 s

Thus,

Time taken to reach 5.0 m above projection point = T-t

=2.85-0.42

2.43 s