Question

A projectile is fired with an initial speed of 300 meters per second at an angle of 45 degrees to the horizontal, neglecting air resistance, find:
A: The time of the impact.
B: The horizontal distance covered (range) in meters and kilometers at the time of the impact.
C: The maximum height in meters of the projectile.

Answer 1

Start by finding the x and y components of the velocity vector.

Answer 2

how do i do that?

Answer 3

Use trigonometry, you need these:

\(\sf sin\theta=\dfrac{opposite}{hypotenuse}\)

\(\sf cos\theta=\dfrac{adjacent}{hypotenuse}\)

Answer 4

ok so theta is equal to 45 degrees, correct?

Answer 5

yes

Answer 6

should my calculator be in radian or degree mode

Answer 7

if you're using degrees (e.g. \(45^o\)) then it should be in degrees

Answer 8

ok, i though so, i just wanted to check though

Answer 9

both cos theta and sin theta were 212.1320344

Answer 10

sounds good. Now we need to use some kinematic equations, do you know them?

Answer 11

these:

\(\sf v_f=v_i+a\Delta t\)

\(\sf x_f=x_i+v_o\Delta t +\dfrac{1}{2}a_x(\Delta t)^2\)

\(\sf v_f^2=v_i^2+2a_x\Delta x\)

Answer 12

wow, um what do i do with those equations

Answer 13

we need to use one (or several) to find what they asked you for, for example:

A: The time of the impact.

The motion is that of a projectile we know that it's an inverted parabola, so the initial and final position (when it touches the ground) are the same. \(\sf y_i=y_f=0\)


so we'd use the second equation: \(\sf y_f=y_i+v_o\Delta t +\dfrac{1}{2}a_y(\Delta t)^2\)

we also know that the only acceleration is due to gravity, a=-9.8 m/\(s^2\), pluggin stuff in gives:

\(\sf 0=0+(212~m/s)\Delta t +\dfrac{1}{2}(-9.8~m/s^2)(\Delta t)^2\)

now we just need to solve for \(\sf \Delta t\).