A diver runs horizontally with a speed of 1.5 m/s off a platform that is 20 m above the water. what is his speed just before hitting the ground?
a ball 1 is thrown horizontally from the top of a building that is 50 meters tall. if another ball 2 is dropped at the same time as the first is released, which ball will hit the ground first?
A. ball 1
B. ball 2
c. they land at the same
D. depends of the initial velocity of ball 1
These two questions are really the same.
in the first the horizontal velocity remains 1.5 the whole time
then vertical problem
20 = 4.9 t^2
solve for t
Thern use that t to get v, speed down at ground
v = g t
v = 9.81 t
then
speed = sqrt (2.25 + v^2)
The second problem is trivial. The horizontal problem has nothing to do with the vertical problem. They fall at the same rate.
I still need help ebcause the answers are
A. 2.5 m/s
B. 5 m/s
C. 1.75 m/s
D. 2 m/s
NOT 19 M/S
A ball rolls off a 1 meters high desk and hits the ground 5 m away from the bottom of the desk. Determine the initial velocity of the ball. Free response
Your problems are all the same.
range = u t
t is time to fall one meter
h = (1/2) g t^2
so
t = sqrt (2 g h) remember that
so
t = sqrt (2*9.81*1)
then
5 meters = u t and we know t
so solve for u
u=5 meters?
To find the diver's speed just before hitting the ground, we can use the principles of motion. The diver experiences two types of motion: horizontal motion (running off the platform) and vertical motion (falling towards the ground).
Let's analyze the vertical motion first. We can use the equation for vertical displacement:
Δy = v_iy * t + (1/2) * g * t^2
Where:
Δy is the vertical displacement (20 m, as given)
v_iy is the initial vertical velocity (0 m/s, as the diver starts from rest)
t is the time taken to reach the ground
g is the acceleration due to gravity (approximately 9.8 m/s^2)
Since the diver starts from rest vertically, the initial vertical velocity v_iy is 0 m/s. Thus, the equation becomes:
20 m = 0 * t + (1/2) * 9.8 m/s^2 * t^2
Now, we can solve for t. Rearranging the equation:
10t^2 = 20
Dividing both sides by 10:
t^2 = 2
Taking the square root of both sides:
t ≈ √2
Hence, t ≈ 1.41 s.
Now, let's analyze the horizontal motion. The diver runs horizontally with a speed of 1.5 m/s. This horizontal speed remains constant throughout the jump. Since there is no acceleration horizontally, we can directly use the speed.
Thus, the diver's speed just before hitting the ground is approximately 1.5 m/s.