Projectile Motion

Question

Dara and Cameron are studying projectile motion in their physics lab class. They set up a Pasco projectile launcher on the edge of their lab table, so that the ball will be launched at an initial height of �= 30.0 inches, initial velocity of �→0= 4.0 m/s and an initial angle of �0=20∘ . They can then record the landing location by placing a piece of carbon paper on the floor some distance away from the launcher. When the ball lands, it will make a mark on the carbon paper.

a) Find horizontal component of initial velocity (two significant figures please). 

b) Find vertical component of initial velocity (two significant figures please).

c) Find the maximum height of the motion (two significant figures please).

d) Find the landing location on carbon paper (three significant figures this time). 

Answer 1

a.) the horizontal component of projectiles launched at an angle will be:

v_x = v₀ cosθ. (Derived from SOH CAH TOA)

Plugging in the values from the problem:

v_x = 4(cos(20º)) = 3.8 m/s

b.) the vertical component of projectiles launched at an angle will be:

v_y = v₀ sinθ. (Derived from SOH CAH TOA)

Plugging in the values:

v_y = 4(sin(20º)) = 1.4 m/s

c.) The maximum height of a projectile occurs when the vertical velocity = 0. The horizontal velocity is constant throughout the motion so be sure to only set v_y=0 at the maximum height.

Using the following kinematic equation we can solve for y (the maximum height).

v_y² = v_y0² + 2a(y – y₀)

Setting v_y = 0, a = –9.8 m/s², and y₀ = 30 inches, but we need to convert that to m.

30 inches * 0.0254 m / 1 inch = 0.762 m

Now plug in our values:

0 = 1.4² + 2(–9.8)(y–0.762)

–1.4² = –2(9.8)(y–0.762)

–1.96 = –19.6(y–0.762)

0.1 = y–0.762

y = 0.86m. So the maximum height is 0.86m above the floor

d.) In order to find the horizontal distance the ball traveled we need to solve for time the launcher is in the air. we will use the following kinematic equation using the vertical values.

y = y₀ +v_y0t + 1/2a_yt²

setting y = 0 because the height of the floor is 0m, y₀ = the height of the desk in m, 0.762 m, and a_y = –9.8 m/s²

Plug in values:

0 = 0.762 + 1.4t +1/2(–9.8)t²

0 = 0.762 + 1.4t – 4.9t²

This is a quadratic and we need to use the quadratic formula to solve.–

Solving for t we get t = –0.2766 and 0.5623 s. Only the positive value makes sense since t must always be positive

Now use the same kinematic equation in the x-direction

x = x₀ +v_x0t + 1/2a_xt²

setting x₀ = 0 and a_x = 0 (the horizontal velocity is constant meaning that a_x = 0 for every projectile motion problem) and t = 0.5623 s

x = 0 + 3.8(0.5623) + 0

x = 2.13 m

Hope this helps!