Question

1.An object falls from rest on a high tower and takes 5.0 s to hit the ground. Calculate the object’s position from the top of the tower at 1.0 s intervals. Make a position-time graph for the object’s motion. In your response, show what you are given, the equation that you used, any algebra required, a table of data, and your graph.
g = 9.8 m/s2

Answer 1

We will be using this equation for this problem
d = ut + ½.at²
Given:
initial velocity, u = 0 (falling from rest) 
acceleration, a = +9.80 m/s²(taking down as the convenient positive direction) 
Time = 1.0s, 2.0s, 3.0s, 4.0s, 5.0s 

Using .. d = ½.at² each time (each calculation is the distance from the top) 
For 1.0s .. d = ½ x 9.80 x (1²) = 4.90 m 
For 2.0s .. d = ½ x 9.80 x (2²) = 19.60 m 
3.0s .. d = 44.10m (you show the working for the rest) 
4.0s .. d = 78.40 m 
5.0s .. d = 122.50m 

Plot distance (displacement from the top) on the y-axis against time on the x-axis (label axes and give units for each).The line of best fit will be a smoothly upward curving line getting progressively steeper. Do not join graph points with straight lines.

Answer 2

Answer:

Position of ball from top of tower after 1 second is 4.9 m

Position of ball: $y=-4.9t^2+122.5$

Step-by-step explanation:

An object falls from rest on a high tower and takes 5.0 s to hit the ground.

• Initial speed, u = 0

• Height of tower, $h_0$

• Acceleration due to gravity, $g=9.8\ m/s^2$

Let position of object from top of tower be y

Using formula, v =u - gt

$v=0-9.8\times 5$

$v=49\ m/s$

Speed of object when hit the ground.

Height of tower, H₀

$H_0=\dfrac{49^2-0}{2\times 9.8}=122.5\ m$

Velocity of ball after 1 s

$v=0-9.8\times 1$

$v=-9.8\ m/s$

Using formula, $s=\dfrac{v^2-u^2}{2g}$

$y=\dfrac{9.8^2-0^2}{2\times 9.8}$

$y=4.9\ m$ From top of tower

Now find position of ball from top of tower.

$y=-\dfrac{1}{2}\times 9.8\times t^2+122.5$

$y=-4.9t^2+122.5$

Please find the attachment for graph.