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:
2.11 seconds
Explanation:
We use the kinematic equation for the velocity in a constantly accelerated motion under the acceleration of gravity (g):
Answer:
Explanation:
The magnitude of the acceleration in the x direction is always zero: TRUE.
At the apex, in the y direction the velocity is zero and the accelration is positive. TRUE.
At the apex, in the x direction, the velocity is zero and the acceleration is zero. FALSE. The accelration is zero, but the velocity is the same it had when it was shot.
The magnitude of the velocity in the y direction is always constant. FALSE, it's subject to gravity and it's velocity varies as
At the apex, in the Y direction, the velocity is negative and the acceleration is zero. FALSE. Velocity is zero, Acceleration is , towards the negative y axis