James asked in Science & MathematicsPhysics · 8 years ago

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?

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• DaveWH
Lv 7
8 years ago

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

• 8 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.

• Jim
Lv 7
8 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