Question

PLEASE HELP!!! WILL MARK BRAINLIEST!! THX! A golfer hits a ball with an initial velocity of 32.7 m/s from the ground. Find the following (round all answers to 2 decimal places...no units):
When does the ball hit the ground? ___

What is the highest height the ball goes? ___

When does the ball reach its max height? ___

How high is the ball after 4.3 seconds? ___

When is the ball 24 meters off the ground? ___ and ___ (enter the smaller value first)

Answer 1

Answer:

See below

Step-by-step explanation:

First Problem

The ball hits the ground when $h(t)=0$, therefore:

$h(t)=-4.9t^2+v_0t+h_0$

$0=-4.9t^2+32.7t$

$0=t(-4.9t+32.7)$

$t=0$ and $t=\frac{32.7}{4.9}\approx6.67$

Since the ball is in the air before it hits the ground, $t=6.67$ (seconds) is the more appropriate choice.

Second Problem

The maximum height of the ball is determined when $t=-\frac{b}{2a}$, therefore:

$t=-\frac{b}{2a}$

$t=-\frac{32.7}{2(-4.9)}$

$t=-\frac{32.7}{-9.8}$

$t\approx3.34$

This means that the height of the ball is at its maximum after 3.34 seconds:

$h(t)=-4.9t^2+32.7t$

$h(3.34)=-4.9(3.34)^2+32.7(3.34)$

$h(3.34)\approx54.55$

Thus, the answer is 54.55 (meters).

Third Problem

Refer to the second problem

Fourth Problem

$h(t)=-4.9t^2+32.7t$

$h(4.3)=-4.9(4.3)^2+32.7(4.3)$

$h(4.3)\approx50.01$

Therefore, the height of the ball after 4.3 seconds is 50.01 (meters).

Fifth Problem

The ball will be 24 meters off the ground when $h(t)=24$, therefore:

$h(t)=-4.9t^2+32.7t$

$24=-4.9t^2+32.7t$

$0=-4.9t^2+32.7t-24$

$t=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$

$t=\frac{-32.7\pm\sqrt{(32.7)^2-4(-4.9)(-24)}}{2(-4.9)}$

$t_1\approx0.84$

$t_2\approx5.83$

Therefore, the ball will be 24 meters off the ground after 0.84 (seconds) and 5.83 (seconds)