Subject:
PhysicsAuthor:
javionpittmanCreated:
1 year agoAnswer:
hii !! I hope this will help u !!!!Author:
popcornfa5t
Rate an answer:
2Given :
Height (H) = 4.9 m
Initial horizontal velocity (u) = 4.5 m/sec
To Find :
Time taken by the stuntman to reach the ground.
Solution :
Velocity of the stuntman in y-axis = ucos∅
= 4.5×cos90
= 0
Therefore, applying the Kinematic equation for uniformly accelerated body,
s = ut + [tex]\frac{1}{2} [/tex]gt²
Where, s = distance covered; u = initial velocity; g = acceleration due to gravity; t = time taken
∴ H = (0)t + [tex]\frac{1}{2} [/tex]gt²
⇒ 4.9 = 0 + [tex]\frac{1}{2} [/tex]×9.8×(t)²
⇒ t² = 1
⇒ t = √1
∴ t = 1 s
The time taken by the stuntman to reach the ground is 1 sec.
Therefore, option (c) is correct
Author:
enrique837
Rate an answer:
17