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
Given:
u(initial velocity)=0
v(final velocity)= 10 m/s
t= 4 sec
Now we know that
v= u + at
Where v is the final velocity
u is the initial velocity
a is the acceleration measured in m/s^2
t is the time measured in sec
10=0+ax4
a=10/4
a=2.5 m/s^2
Option 2 is your answer :)
Hello,
<span>A car with a mass of 2.0×10^3 kg is traveling at 15m/s. We need to find the momentum of the car. To do so, follow this formula:
p=mv
Where,
p = momentum
m = mass
v = </span>velocity
The cars mass is 2.0E3 and its velocity is 15m/s. Therefore:
p=2.0 x 10^3 *15 or 2000(15)
p=30000
Thus, the cars momentum is 30000 kg m/s
Faith xoxo