Imagine an object is thrown upward at 50m/s. At what time will it stop?

Question

Imagine an object is thrown upward at 50m/s. How high will it go? 

Imagine an object is thrown upward at 50m/s. When will it return to zero height?

Answer 1

Background:

This is a standard kinematics question. Recall if you throw an object straight up, at the top of its path the velocity is zero. What we can do is find the time it takes for the object to go to its maximum height and double that for its total time. Remember acceleration due to gravity is always down if we consider up as positive. We will need to solve for time before we can solve for the change in position (height).

Given:

v_i = 50 m/s

v_f = 0 m/s

a_y = - 9.8 m/s²

Δy = ?

t = ?

Equation

v_f = v_i + at

t = (v_f - v_i)/a

Math

t = (v_f - v_i)/a

t = (0 - 50)/(-9.8)

t = 5.1 seconds This means it takes 5.1 seconds to get to the top of its path. We can use this information to solve for the height travelled.

Equation

Δy = v_it + 1/2 at²

Math

Δy = (50)(5.1) + 1/2 (-9.8)(5.1)²

Δy = 127.55 meters

We can also find out the total time in the air by multiplying the time up, by two, as however long it took to get to the highest point, it will take the same amount of time to get back down.

total time in air = 5.1 x 2 = 10.2 seconds