Question

1.) An object is dropped from a height of 200 m. What is its velocity as it reaches the ground? How long time) before it reaches the ground? 2.) An object is fired into a projectile with an initial velocity of 250 m/s at an angle of 60° with respect to the horizontal. What is maximum height it reaches? What is the maximum horizontal distance it reaches? What is the time of flight of the projectile? What is its velocity as it reaches the ground? Locate the object after 6 seconds

Answer 1

h = 200 m

u = 0 m/s

v2 = u + 2gh

2 = 0% + 2 x 9.81 x 200

12 = 3924

V = 62.64 m/s

U=u+ at

62.64 = 0 + 9.81 xt

ta 62.64 9.81 O = 6.385 s

3UUU (2761.561, 2391.582) 2500 2000 1500 1000 -500 (5523.121, 0) 1000 2000 3000 4000 5000 6000 7000

1. Maximum height reached

2 x sine h = 1 29

2502 x sin-60° 2 x 9.81

h=2389.14m

2. Maximum horizontal distance

u2 x sin24 R=-

R= 2502 x sin 120° 9.81

R = 5521 m

3. Time of flight

2u x sing

t =- 2 x 250 x sin 60° 9.81

t = 44.14s

4. Velocity while reaching ground

velocity = ucosi

velocity = 250 x cos60

velocity = 125 m/s

5. Location after 6 seconds

r = utcos

T= 250 X 6 X cos60°

T = 750 m

y = tan .x- 2.42.cos2

9.81 y = tan 60°. 750 - - 2.2502 . cos2 60° 22600 7502

y = 1122.45 m

Please comment for any doubts.