Equations of the vertical launch:
Vf = Vo - gt
y = yo + Vo*t - gt^2 / 2
Here yo = 35.0m
Vo is unknown
y final = 0
t = 4.00 s
and I will approximate g to 10m/s^2
=> 0 = 35.0 + Vo * 4 - 5 * (4.00)^2 => Vo = [-35 + 5*16] / 4 = - 45 / 4 = -11.25 m/s
The negative sign is due to the fact that the initial velocity is upwards and we assumed that the direction downwards was positive when used g = 10m/s^2.
Answer: 11.25 m/s
Answer:
a. 8.96 m/s b. 1.81 m
Explanation:
Here is the complete question.
a) A long jumper leaves the ground at 45° above the horizontal and lands 8.2 m away.
What is her "takeoff" speed v
0
?
b) Now she is out on a hike and comes to the left bank of a river. There is no bridge and the right bank is 10.0 m away horizontally and 2.5 m, vertically below.
If she long jumps from the edge of the left bank at 45° with the speed calculated in part a), how long, or short, of the opposite bank will she land?
a. Since she lands 8.2 m away and leaves at an angle of 45 above the horizontal, this is a case of projectile motion. We calculate the takeoff speed v₀ from R = v₀²sin2θ/g. where R = range = 8.2 m.
So, v₀ = √gR/sin2θ = √9.8 × 8.2/sin(2×45) = √80.36/sin90 = √80.36 = 8.96 m/s.
b. We use R = v₀²sin2θ/g to calculate how long or short of the opposite bank she will land. With v₀ = 8.96 m/s and θ = 45
R = 8.96²sin(2 × 45)/9.8 = 80.2816/9.8 = 8.192 m.
So she land 8.192 m away from her bank. The distance away from the opposite bank she lands is 10 - 8.192 m = 1.808 m ≅ 1.81 m
Answer:
Speed will be equal to 1.40 m/sec
Explanation:
Mass of the rubber ball m = 5.24 kg = 0.00524 kg
Spring is compressed by 5.01 cm
So x = 5.01 cm = 0.0501 m
Spring constant k = 8.08 N/m
Frictional force f = 0.031 N
Distance moved by ball d = 15.8 cm = 0.158 m
Energy gained by spring

Energy lost due to friction

So remained energy to move the ball = 0.0101 - 0.0048 = 0.0052 J
This energy will be kinetic energy


v = 1.40 m/sec
Cold front because the warm front or hit front is lower on the right and goes downwards and cold front is in the middle and goes to Minneapolis
Answer:

Explanation:
The attached figure shows the whole description. Considering the applied force is 100 N.
The acceleration of both blocks A and B, 
Firstly calculating the mass m using the second law of motion as :
F = ma
m is the mass


m = 125 kg
It suddenly encounters a surface that supplies 25.0 N a friction, F' = 25 N



So, the new acceleration of the block is
. Hence, this is the required solution.