Answer:
(a) 62.5 m
(b) 7.14 s
Explanation:
initial speed, u = 35 m/s
g = 9.8 m/s^2
(a) Let the rocket raises upto height h and at maximum height the speed is zero.
Use third equation of motion
h = 62.5 m
Thus, the rocket goes upto a height of 62.5 m.
(b) Let the rocket takes time t to reach to maximum height.
By use of first equation of motion
v = u + at
0 = 35 - 9.8 t
t = 3.57 s
The total time spent by the rocket in air = 2 t = 2 x 3.57 = 7.14 second.
Answer:
D. from a separate pool than is the control group.
Explanation:
in the picture the person answers is backwards but...
hope this helps have a nice day
Answer:
a
The height is
b
The horizontal distance is
Explanation:
From the question we are told that
The speed is
The angle is
The height of the cannon from the ground is h = 2 m
The distance of the net from the ground is k = 1 m
Generally the maximum height she reaches is mathematically represented as
=>
=>
Generally from kinematic equation
Here s is the displacement which is mathematically represented as
s = [-(h-k)]
=> s = -(2-1)
=> s = -1 m
There reason why s = -1 m is because upward motion canceled the downward motion remaining only the distance of the net from the ground which was covered during the first half but not covered during the second half
a = -g = -9.8
So
=>
using quadratic formula to solve the equation we have
Generally distance covered along the horizontal is
=>
=>