# Projectiles Problem: Time to Hit Ground from 45m

• MHB
• omartharwat
In summary, a ball is projected vertically from the top of a building with a height of 45m from Earth's surface. Its velocity is determined by the relation v= 40-10t. To find the time it takes for the ball to hit the ground, we use the equation y(t) - y(0) = 40 t - 5t^2 or y(t) - y(0) = v_0 t + 1/2 a t^2. After solving for t, we get t= 9 seconds, which means the ball will hit the ground 9 seconds after being thrown upward.
omartharwat
a ball is projected vitically from the top of a building with hight of 45 m from Erath`s surface . it velocity is determind by the relation v= 40-10t how much time the ball will take to hit the ground?

Well, we need an equation relating the velocity with the distance traveled. We can integrate:
$$\displaystyle \int_0^t v(t) ~ dt = \int_0^t (40 - t) ~ dt \implies y(t) - y(0) = 40 t - 5t^2$$

If you can't use Calculus then consider that the acceleration ($$\displaystyle a = \Delta v / \Delta t$$) is constant. So we can use y(t) - y(0) = v_0 t + 1/2 a t^2[/math], which gives the same thing.

Give it a try and show us what you get.

-Dan

There is a constant acceleration, $-g= -9.8 m/s^2$ so after time t, the speed is $v(t)= v_0- 9.8t m/s$ where $v_0$ is the initial speed. The problem says that the velocity is $40- 10t$. I suspect that "10" is an approximation to g= 9.8 so that $v_0= 40 m/s$.

The height after t seconds will be $h(t)= 45+ 40t- 5t^2$ (the "45" is the initial height at t= 0).

The "ground" is at h= 0 so to find the time until the ball hits the ground you need to solve the equation $h(t)= 45+ 40t- 5t^2= 0$ for t.

First divide both sides by -5 to get $t^2- 8t- 9= (t- 9)(t+ 1)= 0$. The roots are t= 9 and t= -1. Since the ball cannot hit the ground before it was thrown up, t= 9. The ball hits the ground 9 seconds after being thrown upward.

(t= -1 can be interpreted as when the ball could have been thrown up from the ground to be at 45 feet at t= 0.)

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## What is a projectile?

A projectile is any object that is thrown or launched into the air and is subject to the force of gravity. Examples of projectiles include a baseball, a bullet, or a rocket.

## What is the equation for calculating the time it takes for a projectile to hit the ground?

The equation for calculating the time it takes for a projectile to hit the ground is t = √(2h/g), where t is the time in seconds, h is the initial height in meters, and g is the acceleration due to gravity (9.8 m/s²).

## How do you solve a projectile problem?

To solve a projectile problem, you need to first identify the given information, such as the initial height, launch angle, and initial velocity. Then, use the equation t = √(2h/g) to calculate the time it takes for the projectile to hit the ground. Finally, use the equation d = v0t + ½at² to calculate the horizontal distance traveled by the projectile.

## What factors affect the time it takes for a projectile to hit the ground?

The time it takes for a projectile to hit the ground is affected by the initial height, launch angle, and initial velocity. Other factors that may affect the time include air resistance and wind speed.

## Can the time it takes for a projectile to hit the ground be negative?

No, the time it takes for a projectile to hit the ground cannot be negative. Time is always a positive value and represents the duration of an event. If the calculated time is negative, it means that the projectile has not yet been launched or has already hit the ground.

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