Question

A ball rolls off a tall shelf with a horizontal speed of 2 m/s. The shelf is 1.29 m high. How long will it take for the ball to strike the ground?

Answer 1

consider the motion of the ball in vertical direction after it rolls off the shelf

v₀ = initial velocity in vertical direction just after it leaves shelf = 0 m/s

Y = vertical displacement = height of the shelf = 1.29 m

a = acceleration of ball = acceleration due to gravity = 9.8 m/s²

t = time to strike the ground

Using the kinematics equation

Y = v₀ t + (0.5) a t²

1.29 = (0) t + (0.5) (9.8) t²

1.29 = 4.9  t²

t² = 1.29/4.9

t² = 0.2633

t = √ 0.2633

t = 0.51 sec

hence the ball strike the ground after 0.51 sec

Answer 2

The time it will take the ball to strike the ground is 0.51seconds

Using Newton's law of motion. The required formula will be:

$S =ut+\frac{1}{2}at^2$

Since the acceleration acting on the object is the acceleration due to gravity, hence;

$S =ut+\frac{1}{2}gt^2$

Given the following parameters

Horizontal speed u = 2m/s

Height S = 1.29m

g is the acceleration due to gravity= 9.8m/s²

Substitute the given parameters into the expression to have:

$1.29=2t+\frac{1}{2}*9.8t^2\\1.29=2t+4.9t^2\\4.9t^2+2t-1.29=0$

Factoring the result will give t = 0.51secs

Hence the time it will take the ball to strike the ground is 0.51seconds

Learn more here: https://brainly.com/question/8898885