Physics motion.

Question

Throw the coin on a plate that is higher than the hand and far from the hand 2.1m.

If the coin is thrown into the plate by v=6.4m/s at an angle of 60 degrees

A) the height of the plate compared to the hand

B) v in rectilinear motion of the coin before it on a plate

Answer 1

Hi Kittipat P.,

To find the distance of the plate above the hand we can use the equation y = y_o + v_oy*t + 1/2 at². If we choose y to be positive in the up direction (a = -g), the hand that flips the coin to be the initial height of zero (y₀ = 0), and the initial velocity in the y direction is the initial velocity given times the sine of the angle, v_oy = v_o*cos(θ). We end up with y = v_o*sin(θ)*t - 1/2gt². We know v_o (6.4 m/s), θ (60^o), and g (9.8 m/s²), but we need to know time t in order to solve for y.

We can find the time it takes the coin to get to the plate in the x direction (which is the same time it takes in the y direction), by using x = v_ox*t, and v_ox = v_o*cos(θ).

We have:

x = v_o*cos(θ)*t

2.1 m = 6.4 m/s * cos(60^o) * t

2.1 m /(6.4 m/s *cos(60^o) = t

0.67 s = t

Now solve for y using time t:

y = v_o*sin(θ)*t - 1/2gt²

y = 6.4 m/s * sin(60^o) * 0.67 - 4.8 * 0.67²

y = 1.5 m above the hand

Have fun! I hope this helps, Joe.