Answer:
28.5 m/s
18.22 m/s
Explanation:
h = 20 m, R = 20 m, theta = 53 degree
Let the speed of throwing is u and the speed with which it strikes the ground is v.
Horizontal distance, R = horizontal velocity x time
Let t be the time taken
20 = u Cos 53 x t
u t = 20/0.6 = 33.33 ..... (1)
Now use second equation of motion in vertical direction
h = u Sin 53 t - 1/2 g t^2
20 = 33.33 x 0.8 - 4.9 t^2 (ut = 33.33 from equation 1)
t = 1.17 s
Put in equation (1)
u = 33.33 / 1.17 = 28.5 m/s
Let v be the velocity just before striking the ground
vx = u Cos 53 = 28.5 x 0.6 = 17.15 m/s
vy = uSin 53 - 9.8 x 1.17
vy = 28.5 x 0.8 - 16.66
vy = 6.14 m/s
v^2 = vx^2 + vy^2 = 17.15^2 + 6.14^2
v = 18.22 m/s
Answer:
Im actually depressed
Explanation:
jk thx for the points thoo
Answer:
25J
Explanation:
Given parameters:
Mass of the dog = 10kg
Speed of the dog = 5m/s
Unknown:
The minimum energy required to stop the dog = ?
Solution:
The dog is moving with a kinetic energy and to stop the dog, an equal amount of kinetic energy generated must be applied to the dog.
To find the kinetic energy;
K.E =
m v²
m is the mass
v is the velocity
Now insert the parameters and solve;
K.E =
x 10 x 5 = 25J
Step 1 : Get your supply list together
Step 2 : Pick what model you want to do
Step 3 : Ask for a partner
Step 4 : Complete the model and take your time.
Step 5 : Read the directions carefully