Calculus application 1. Supposed that a ball is dropped from the upper observation deck of the Horizons 101 Tower 1., located
178 m above the ground. (a) What is the velocity of the ball after 5 sec? (b) How fast is the ball traveling
when it hits the ground?

49m/s

59.07 m/s

Step-by-step explanation:

Given that :

Distance (s) = 178 m

Acceleration due to gravity (a) = g(downward) = 9.8m/s²

Velocity (V) after 5 seconds ;

The initial velocity (u) = 0

Using the relation :

v = u + at

Where ; t = Time = 5 seconds ; a = 9.8m/s²

v = 0 + 9.8(5)

v = 0 + 49

V = 49 m/s

Hence, velocity after 5 seconds = 49m/s

b) How fast is the ball traveling when it hits the ground?

V² = u² + 2as

Where s = height = 178m

V² = 0 + 2(9.8)(178)

V² = 0 + 3488.8

V² = 3488.8

V = √3488.8

V = 59.07 m/s

lolololkzaxio2ier286udeg