Question

A ball is thrown horizontally from the top of a building 37.5 m high. The ball strikes the ground at a point 80.3 m from the base of the building. The acceleration of gravity is 9.8 m/s 2 . Find the time the ball is in motion. Answer in units of s.

Answer 1

Answer:

t = 2.77 s

Explanation:

The ball in its movement describes a curved line called a semiparabola, therefore two coordinates are required to fix the position at each instant of time, since the movement is performed in the X-Y plane:

Equation of movement of the ball in the X axis

X = v₀x*t   Equation  (1)

Equation of movement of the ball in the Y axis

Y = y₀+ v₀y*t -½ g*t² Equation  (2)

Where

X : horizontal position in meters (m)  

Y : vertical  position in meters (m)  

y₀ : initial vertical  position in meters (m)

v₀x :  X-initial speed in m/s  

v₀y :  Y-initial speed in m/s  

g: acceleration due to gravity in m/s²

t : time to position (X,Y)

Data

y₀ = 37.5 m

v₀y = 0

g = 9.8 m/s²

Problem development

The time the ball remains in the air is the same as the ball takes to touch the floor, that is, Y = 0

We apply the Equation (2):

Y = y₀+ (v₀y)*t - (½) g*t²

0 =  37.5 +(0)*t- (1/2)*(g)*t²

0 =  37.5 - (1/2)*(9.8)*t²

(1/2)*(9.8)*t²  = 37.5

t² = (2)(37.5)/(9.8)

$t= \sqrt{\frac{2*37.5}{9.8} }$

t = 2.77 s