Answer:20.82 m
Explanation:
Given
distance between wall and launcher=5 m
initial velocity=30 m/s
Launching angle
Height of wall=12 m
maximum height by ball
h=22.95 m
y=4.72 m
so ball will strike with wall
and for perfect restitution final velocity of ball will be same as the horizontal velocity before its impact, only direction will be opposite and vertical velocity will be zero
thus it seems as if someone throw the ball with horizontal velocity of 30cos45 from a height of 4.72
Time required to cover 4.72 m
t=0.981 s
Horizontal distance traveled in this time
R=20.82 m
Answer:
yes because there is also a theory about this I studied in biology
theory name -Lamarck's theory
Answer:
U = 30 m/s
a = 3 m/s²
Explanation:
"A car accelerating uniformly from rest reaches a maximum speed of U in 10 s. It then moves with that speed for an additional 20 s. The distance covered by the car in the 30 s interval is 750 m. Find U and the acceleration of the car in the first 10 s."
During the first 10 s:
v₀ = 0 m/s
v = U m/s
t = 10 s
The distance covered in this time is the average velocity times time:
Δx = ½ (v + v₀) t
Δx = ½ (U + 0) (10)
Δx = 5U
The distance covered in the next 20 seconds is speed times time:
Δx = 20U
The total distance is 750 m:
5U + 20U = 750
25U = 750
U = 30 m/s
The acceleration during the first 10 seconds is the change in speed over change in time.
a = Δv / Δt
a = (30 m/s − 0 m/s) / 10 s
a = 3 m/s²
Answer:
1.5
x <u>1</u><u>.</u><u>0</u>
1.50
x <u>0</u><u>.</u><u>5</u>
075.00
answer: 75.00m
Explanation:
I hope this help
