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?
- 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..
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
- 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