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2 votes
CAN SOMEONE PLEASE HELP ME its way to confusing for me so i will give brainy and full stars to as long as its with both questions

Question 4.

A cannonball is shot from a cannon into the air from a height of 50m with an initial velocity of 75m and s. After how many seconds does the cannonball hit the ground?

CAN SOMEONE PLEASE HELP ME its way to confusing for me so i will give brainy and full-example-1
asked
User Myrline
by
7.7k points

1 Answer

3 votes

for the cannon ball one

I can use the equation of motion to determine the time taken by the cannonball to hit the ground:

h(t) = -0.5gt^2 + V₀t + h₀

where:

- g is the acceleration due to gravity, -9.8 m/s^2

- V₀ is the initial velocity, 75 m/s

- h₀ is the initial height, 50 m

I'll set h(t) to 0, since the ball will hit the ground when its height is 0, and solve for t:

0 = -0.5(-9.8)t^2 + 75t + 50

0 = 4.9t^2 + 75t + 50

Using the quadratic formula:

t = (-b ± √(b^2 - 4ac)) / 2a

where a = 4.9, b = 75, and c = 50.

t = (-75 ± √(75^2 - 4(4.9)(50))) / 2(4.9)

t = (-75 ± √(5625 - 980)) / 9.8

t = (-75 ± √4645) / 9.8

t ≈ 7.55 seconds (rounding to 2 decimal places)

Therefore, the cannonball will hit the ground after approximately 7.55 seconds.

for the second one

since I don't have any specific values for g, v₀, or h, I can still write the general form of the projectile motion equation:

h(t) = -0.5gt^2 + v₀t + h₀

where g is the acceleration due to gravity, v₀ is the initial velocity, and h₀ is the initial height.

I can then plug in these variable values into Desmos to plot the trajectory of the projectile.

answered
User Keeleon
by
8.6k points
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