Projectile Motion problem help please

[Newell33]

Homework Statement

Use 4 decimal places at each step to ensure accuracy.
a) A rocket is launched vertically and remains in the air for 5.10 seconds until it hits the ground. Calculate the initial velocity of the rocket. Calculate the maximum height of the rocket (positive).
vo = 24.99

m/s (Note: You will need this initial velocity for parts b, c, and d)
y = 31.8622

m

b) The rocket is now launched with the same initial velocity as calculated in part A at an angle of 45 degrees. Calculate the time the rocket will be in the air for. Calculate the maximum height that the rocket will reach. Calculate the horizontal distance that the rocket will travel until it hits the ground.
t = s

y = m
x = m

Homework Equations

idk sorry :(

The Attempt at a Solution

ok so i tried to do
t = (v - vo) / -9.8


and its wrong because i got
2.55

 

[PhizKid]

If you don't know the kinematic equations, you won't be able to do solve these questions unless you know how to derive them yourself. You can find some of the formulas (not all of them, but you can derive the rest from the basic ones) in the sticky for this forum: https://www.physicsforums.com/showthread.php?t=110015

 

[Spinnor]

Do you have a physics textbook? If yes which one is it? I think I got through intro physics by studying the worked examples in my physics book (and also reading the book).

 

[Basic_Physics]

This: t = (v - vo) / -9.8
gives you only half of the time because its speed is (momentarily) zero at the top.
Its speed is not zero when it hits the ground!

 

[Nickie]

s for the time and 31.86 m for the height.

Dear student,

To solve this projectile motion problem, we can use the following equations:

1. For vertical motion:

y = yo + voy*t - 1/2*g*t^2

where y is the final height, yo is the initial height, voy is the initial vertical velocity, t is the time, and g is the acceleration due to gravity (-9.8 m/s^2).

2. For horizontal motion:

x = xo + vox*t

where x is the final horizontal distance, xo is the initial horizontal distance (which can be assumed to be 0 in this case), vox is the initial horizontal velocity, and t is the time.

a) To find the initial velocity (vo) of the rocket, we can use the equation for vertical motion with the given values:

y = 0 (since the rocket starts and ends at ground level)
yo = 0 (since the rocket starts at ground level)
t = 5.10 s (given in the problem)
g = -9.8 m/s^2 (acceleration due to gravity)

Plugging in these values and solving for voy, we get:

0 = 0 + voy*5.10 - 1/2*(-9.8)*(5.10)^2

voy = 24.99 m/s

Therefore, the initial velocity of the rocket is 24.99 m/s.

To find the maximum height of the rocket, we can use the same equation for vertical motion with the given values:

y = maximum height (since we want to find the maximum height)
yo = 0 (since the rocket starts at ground level)
voy = 24.99 m/s (calculated in the previous step)
t = 5.10 s (given in the problem)
g = -9.8 m/s^2 (acceleration due to gravity)

Plugging in these values and solving for y, we get:

y = 0 + 24.99*5.10 - 1/2*(-9.8)*(5.10)^2

y = 31.8622 m

Therefore, the maximum height of the rocket is 31.8622 m.

b) For the rocket launched at an angle of 45 degrees, we need to consider both vertical and horizontal motion.

To find the time the rocket will be in the air for