Answer:
Height is 11.25m
Explanation:
<u>Given the following data;</u>
Initial velocity, u = 0
Final velocity, v = 15m/s
Acceleration due to gravity, g = 10m/s²
To find the height, we would use the third equation of motion;
Where;
- V represents the final velocity measured in meter per seconds.
- U represents the initial velocity measured in meter per seconds.
- a represents acceleration measured in meters per seconds square.
- S represents the displacement (height) measured in meters.
<em>Making S the subject, we have;</em>

But a = g = 10m/s²
<em>Substituting into the equation, we have;</em>

S = 11.25m
<em>Therefore, the ball will reach a height of 11.25m before it begins to fall. </em>
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:
Air resistance/Drag (this is the proper term for air resistance) or gravity or friction
It depends on the direction and if you throw it across a surface or if you throw it straight upwards.
I hope this helps!!
Answer:
3.25 × 10^7 m/s
Explanation:
Assuming the electrons start from rest, their final kinetic energy is equal to the electric potential energy lost while moving through the potential difference (ΔV)
Ek = 1/2 mv2 = qΔV .................. 1
Given that V is the electron speed in m/s
Charge of electron = 1.60217662 × 10-19 coulombs
Mass of electron = 9.109×10−31 kilograms
ΔV = 3.0kV = 3000V
Make V the subject of the formula in eqaution 1
V = sqr root 2qΔV/m
V = 2 × 1.60217662 × 10-19 × 3000 / 9.109×10−31
V = 3.25 × 10^7 m/s
Explanation:
formula for energy is k. e = ½mv²
m= 9
v= 75
k. e = ½×9×75 =337•5