Will someone please help me with these Physics questions

Question

Now that you know this follow this steps to solve the problem: A cannonball is shot out of a cannon at 0º to the horizontal of a cliff with an initial speed of 25.0 m/s from a height of 40 m. How far away from the cliff does the cannonball land?

a. 25 m

b. 229 m

c. 1125 m

d. 71.3 m

Reading the data provided, does this object move at a constant rate? 

a. Yes, because the best fit line matches a linear relationship.

b. No, because the dots are not all on the line.

c. Yes, the best fit line is exponent.

d. No, because the best fit line should be quadratic.

Use the following pictures to help you match the picture of the graph of motion with the current occurrence. 

 

A student has a red Pasco smart cart that graphs data of the motion of the cart on their computer. They are trying different motions to explore how motion is graphed in 1 dimension. Here is their first attempt. 

 

Occurence #1

The student starts at the origin and moves away at a steady pace. Then, they pause the cart for a few moments and return the cart back to the origin. 

 

Graph #1

Graph #2

Graph #3

Graph #4

a. Graph #2

b. Graph #1

c. Graph #3

d. Graph #4

Answer 1

Solution:

You are given initial velocity V₀ = 25 m/s , initial height y₀ = 40 m/s and the angle ∝ =0 and need to find the range x of a projectile shot at 0 degrees, x = V_o cos ∝ t , x = 25 m/s cos (0º) t = 25m/s (t)

so you need to find the time, from the height equation of a projectile

y = -1/2 gt² + V₀ sin ∝ t + y_o , knowing that the final height of the cannon shot is zero, y = 0

0 = -1/2 gt² + V_o sin ( 0) + 40 m , 0 = -1/2 gt² + 40m, 1/2 gt² = 40 m , 1/2(9.8 m/.s²) t² = 40 m

t² = (40m)(2) / 9.8 m/s² = 8.16 s² and t = √ 8.16 s² = 2.85 s

and x = 25m/s ( 2.85 s) = 71.3 m Answer D