Worksheet 2: Projectile motion
1) A projectile is shot at time t = 0 from the
edge of a cliff 115 m above ground level
with an initial speed of 65.0 m/s at an angle
of 35.0° with the horizontal, as shown in the
Figure.
xmax, ymax
x0, y0
+y
+x
(a) Would a projectile shot from the same spot on the cliff with the same initial y
velocity, v0y, but smaller x velocity, v0x, hit the ground earlier, later or at the same
time?
At the same time. Only the y component of the velocity (v0y) matters for the
change of the y coordinate.
(b) Give the y component of the velocity (v0y) immediately after the projectile is shot.
υ0 = 65.0 m/s, θ 0 = 35.0°, and therefore v y0 = v0 sin θ0 = (65.0 m/s) sin 35.0° = 37.3 m/s
(c) Determine the time t taken by the projectile to hit point P at ground level. (Hint:
you will get a quadratic equation in t, which you need to solve.)
Choose the origin to be at ground level, under the place where the projectile is
launched, and upward to be the positive y direction. For the projectile,
a y = −g; y = 0; y0 = h.
The time it takes to reach the ground is found from the equation for the y
position, with a final height of 0.
y = y0 + υ y0t + 12 a y t 2
t=
→
0 = y0 + υ0 sin θ 0t − 12 gt 2
−υ0 sin θ0 ± υ02 sin 2 θ0 − 4 − 12 g h
(
2
(
− 12 g
)
)
→
= 9.964 s, −2.3655 s = 9.96 s
Choose the positive time since the projectile was launched at time t = 0.
Worksheet 2: One-dimensional motion and relative motion
(d) Determine the distance X of point P from the base of the vertical cliff.
The horizontal range is found from the horizontal motion at constant velocity
using the calculated time t from part (b).
Δx = υ x t = (υ0 cos θ 0 )t = (65.0 m/s)(cos 35.0°)(9.964 s) = 531 m
(e) Calculate the horizontal and the vertical components of its velocity at the instant
just before the projectile hits point P.
At the instant just before the particle reaches the ground, the horizontal
component of its velocity is the constant
v x = v0 cos θ0 = (65.0 m/s) cos 35.0° = 53.2 m/s .
The vertical component is found from the equation for the velocity as function
of time:
υ y = υ y0 + at = υ0 sin θ 0 − gt = (65.0 m/s) sin 35.0° − (9.80 m/s2 )(9.964 s)
= −60.4 m/s
(f) Calculate the magnitude of the velocity at the instant just before the projectile hits
point P.
The magnitude of the velocity is found from the x and y components
calculated in part (c) above.
υ = υ x2 + υ 2y = (53.2 m/s)2 + (−60.4 m/s)2 = 80.5 m/s
(g) Calculate the angle made by the velocity vector with the horizontal at the instant
just before the projectile hits point P.
The direction of the velocity is θ = tan −1
is moving 48.6° below the horizontal .
υy
υx
⎛ −60.4 ⎞
= tan −1 ⎜
= −48.6°,
⎝ 53.2 ⎟⎠
so the object
Worksheet 2: One-dimensional motion and relative motion
(h) Find the maximum height above the cliff top reached by the projectile.
The maximum height above the cliff top reached by the projectile will occur
when the y velocity is 0 and is found from the equation relating velocity
(squared), acceleration and displacement.
υ 2y = υ 2y0 + 2a y ( y − y0 ) → 0 = υ02 sin 2 θ0 − 2g( ymax − y0 )
ymax − y0 =
υ02 sin 2 θ0 (65.0 m/s)2 sin 2 35.0°
=
= 70.9 m
2g
2(9.80 m/s2 )
(i) At what time does the projectile reach the highest point in its trajectory?
υ y = υ0 y + a y tmax =υ0 sin θ + a y tmax → tmax =
υ y − υ0 sin θ
ay
=
−υ0 sin θ
( −g )
= 3.80 s
(j) How far from the launch point is the highest point of the trajectory?
The distance along the x axis between the launch point is the highest point of
the trajectory is given by
xmax − x0 = υ x tmax = (υ0 cos θ0 )tmax = (65.0 m/s)(cos 35.0°)(3.80 s) = 202 m
Using ymax from part (g) the distance between the launch point and the highest
point of the trajectory is
dmax =
(x
max
− x0
2
) +(y
max
− y0
)
2
= 214 m
Answers:
1.b) 37.3 m/s c) 9.96 s d) 531 m e) 53.2 m/s; −60.4 m/s f) 80.5 m/s g) −48.6o h) 70.9 m i) 3.80 s j) 214 m