# Please Help! Projectile Motion Question. Thanks!?

A diver running 2.1 m/s dives out horizontally from the edge of a vertical cliff and 2.6 s later reaches the water below. 1. How high was the cliff? 2. How far from its base did the diver hit the water?

### 4 Answers

- DaveWHLv 78 years agoFavorite Answer
Try to spend a little time memorising the equations of motion, and these things become far less daunting. From those equations there is one that looks like this..

1)

For an object falling under gravity, g is positive, so

Vertical height = ut + (1/2) gt^2

where u = initial vertical velocity; t = time of fall; g = acceleration due to gravity.

Here, there is no initial velocity in the vertical direction so

Vertical height = (1/2) g t^2

Height = 0.5 * 9.8 * 2.6^2 = 33.1 m

2) His horizontal velocity does not change throughout the flight.

Distance = Velocity * time = 2.1 * 2.6 = 5.46 m

- 8 years ago
first ill answer the second part....the distance from the base will be = horizontal velocity X time of flight

= 2.1 X 2.6

= 5.46 m

for the first part you must consider the formula s= ut + 1/2 g (t) square in vertical direction

since the velocity is horizontal so vertical velocity will be zero so what remains is

s= 1/2 g (t)square

=1/2 X 9.8 X 2.6 X 2.6

=33.124 m

- oldprofLv 78 years ago
S = 1/2 gT^2 = 4.9*2.6^2 = 33.124 m. ANS 1.

X(T) = Ux T = 2.1*2.6 = 5.46 m ANS 2.

- JimLv 78 years ago
Free-Fall time determines height, by h = 1/2gt² = (0.5)(9.8)(2.6)² = 33 m ANS-1

range = (2.1)(2.6) = 5.5 m ANS-2