A projectile, fired with unknown initial velocity, lands 18.4 s later on the side of a hill, 3290 m away horizontally and 409 m vertically above its starting point. (Ignore any effects due to air resistance.)
(a) What is the vertical component of its initial velocity?
112.38
(b) What is the horizontal component of its initial velocity?
178.8
(c) What was its maximum height above its launch point?
(d) As it hit the hill, what speed did it have and what angle did its velocity make with the vertical?
maximum height is when vertical velocity is zero, solve for that time, then find h.
for d) I would find vertical velocity when it hit, horizonal velocity, and add them as a vector. The angle can be found frm these two components.
To find the answers to these questions, we can break down the projectile motion into its horizontal and vertical components. Let's go step by step:
(a) To find the vertical component of the initial velocity, we need to use the equation of motion for the vertical direction. The equation is given by:
y = u_y * t + (1/2) * g * t^2
where:
y is the vertical displacement (409 m),
u_y is the vertical component of the initial velocity (which we need to find),
t is the time of flight (18.4 s), and
g is the acceleration due to gravity (-9.8 m/s^2).
Plugging in the given values, we can rearrange the equation to solve for u_y:
409 = u_y * 18.4 + (1/2) * (-9.8) * (18.4)^2
Solving this equation will give us the value of u_y. Let's calculate it:
u_y = (409 - (1/2) * (-9.8) * (18.4)^2) / 18.4
Calculating this expression will give us the vertical component of the initial velocity as 112.38 m/s.
(b) To find the horizontal component of the initial velocity, we can use the equation that relates horizontal displacement, initial horizontal velocity, and time of flight:
x = u_x * t
where:
x is the horizontal displacement (3290 m),
u_x is the horizontal component of the initial velocity (which we need to find), and
t is the time of flight (18.4 s).
Plugging in the given values, we can solve for u_x:
3290 = u_x * 18.4
Solving this equation will give us the value of u_x. Let's calculate it:
u_x = 3290 / 18.4
Calculating this expression will give us the horizontal component of the initial velocity as 178.8 m/s.
(c) To find the maximum height above the launch point, we need to consider the vertical motion. At the maximum height, the vertical component of the velocity will be zero. We can use the equation of motion for the vertical direction to find this height:
0 = u_y + g * t_max
where:
u_y is the vertical component of the initial velocity (which we found in part a),
t_max is the time at which the projectile reaches its maximum height, and
g is the acceleration due to gravity (-9.8 m/s^2).
Solving this equation will give us the value of t_max. Let's calculate it:
t_max = -u_y / g
Substituting the value of u_y obtained in part a, we have:
t_max = -112.38 / -9.8
Calculating this expression will give us the time at which the projectile reaches its maximum height. We can then use this time to find the vertical displacement at that point (the maximum height).
(d) To find the speed and angle of the velocity vector at the moment the projectile hits the hill, we can use the equations for speed and angle:
speed = sqrt(u_x^2 + u_y^2)
angle = arctan(u_y / u_x)
Substituting the values of u_x and u_y obtained in parts (a) and (b), respectively, we can calculate the speed and angle.
Note: Values could differ slightly when considering significant figures and rounding off numbers.