Projectile Review Worksheet

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Table of Contents

Problems

1. Marlene jumps off the edge of a cliff and hits the water 1.5 seconds later, about 4.5 m from the base of the cliff. What height was the cliff? With what speed did she leave the edge?

So we fill in our H/V suvat:

H V

s = 4.5 m
u =
v =
a = 0
t = 1.5 s

s =
u = 0 (cliff)
v =
a = -9.81 m/s/s
t = 1.5 s

We are looking for vertical displacement (s) and horizontal velocity (u). We can solve for each variable separately. For vertical displacement, we can use the formula

s = ut + 1/2at²

s = (0 m/s)(1.5 s) + (.5)(-9.81 m/s²)(1.5 s)²

s = 11.03625 m = 11 m

For horizontal velocity, we can use the formula

u = s/t

u = (4.5 m) / (1.5 s)

u = 3 m/s

Table of Contents

2. Question

H V

s = 3.4 m
u =
v =
a = 0
t =

s = 21
u = 0 (cliff)
v =
a = -9.81 m/s/s
t =

The first thing we are going to solve for is time (t). We will do this using the vertical side of the table.

s = ut + 1/2at²
21 m = (0 m/s)(t) + (.5)(9.81 m/s²)(t²)
21 m = 0 + (4.905)(t²)
t = 2.0691 s = 2.07 s

With our newfound value of time, we can now solve for the horizontal velocity (u).

u = s/t
u = (3.4 m) / (2.07 s)
u = 1.6425 m/s = 1.64 m/s