A T-shirt cannon mounted at the top of an arena needs to fire a t-shirt into the first row, a horizontal distance of 39 meters away. If the cannon launches t-shirts at 12 m/s, how high is the cannon mounted? Question 1 options:
3.3 m
16.2 m
53.4 m
8.9 m

h = 51.75 m

nearest answer is:

53.4 m

Explanation:

First we analyze the horizontal motion. Since, the air friction is assumed to be negligible. Hence, the horizontal motion shall be uniform. Therefore,

s = V₀ₓ t

where,

s = horizontal distance = 39 m

V₀ₓ = Horizontal Initial Velocity = 12 m/s

t = time = ?

Therefore,

39 m = (12 m/s)t

t = 39 m/12 m/s

t = 3.25 s

Now, we analyze the vertical motion. Applying newton's second equation of motion to vertical motion:

h = V₀y t + (1/2)gt²

where,

h = height of cannon = ?

V₀y = initial vertical velocity = 0 m/s

g = 9.8 m/s²

Therefore,

h = (0 m/s)(3.25 s) + (1/2)(9.8 m/s²)(3.25 s)²

h = 51.75 m

a lot of it means it can be "burned" for energy and sold at one price. if it is scarce, the price will go up

massxgravity x heght=weightxheight= pe

4900=490xh

h=10