Answer:
find the diagram in the attachment.
Explanation:
Let vi = 12 m/s be the intial velocy when the ball is thrown, Δy be the displacement of the ball to a point where it starts returning down, g = 9.8 m/s^2 be the balls acceleration due to gravity.
considering the motion when the ball thrown straight up, we know that the ball will come to a stop and return downwards, so:
(vf)^2 = (vi)^2 + 2×g×Δy
vf = 0 m/s, at the highest point in the upward motion, then:
0 = (vi)^2 + 2×g×Δy
-(vi)^2 = 2×g×Δy
Δy = [-(vi)^2]/2×g
Δy = [-(-12)^2]/(2×9.8)
Δy = - 7.35 m
then from the highest point in the straight up motion, the ball will go back down and attain the speed of 12 m/s at the same level as it was first thrown
Answer:
Explanation:
a) Using the equation of motion
S = ut + 1/2gt²
S is the distance of fall
g is the acceleration due to gravity
t is the time taken
Given S = 12.0m, g = 9.81m/s^2, un= 0m/s
12 = 0+1/2(9.81)t²
12 = 4.905t²²²
t² = 12/4.905
t² = 2.446
t = √2.446
t = 1.56secs
b) To determine how fast is the frog falling at this point, we need to calculate the speed of the frog. Using the equaton v = u+gt
v = 0+9.81(1.56)
v = 15.34m/s
Hence the frog is falling at the rate of 15.34m/s
D............................
Answer:
you need to consider the use for the product, how brittle the materials are, how they react to certain things, the cost of the materials, the durability and flexibility of the materials, and how easy to obtain the materials are as well as how they would work and how they would hold
Answer:
0.000025s
Explanation:
Period it’s. : T(s)= 1/f(Hz)=1/40000Hz=0.000025s
