Answer:
15.7 m/s
Explanation:
The motion of the cannonball is a accelerated motion with constant acceleration g = 9.8 m/s^2 towards the ground (gravitational acceleration). Therefore, the velocity of the ball at time t is given by:
where
u = 0 is the initial velocity
g = 9.8 m/s^2 is the acceleration
t is the time
If we substitute t=1.6 s into the equation, we find the final velocity of the cannonball:
Answer:
Time period of the motion will remain the same while the amplitude of the motion will change
Explanation:
As we know that time period of oscillation of spring block system is given as
now we know that
M = mass of the object
k = spring constant
So here we know that the time period is independent of the gravity
while the maximum displacement of the spring from its mean position will depends on the gravity as
so we can say that
Time period of the motion will remain the same while the amplitude of the motion will change
Answer:
(a). The initial velocity is 28.58m/s
(b). The speed when touching the ground is 33.3m/s.
Explanation:
The equations governing the position of the projectile are
where is the initial velocity.
(a).
When the projectile hits the 50m mark, ; therefore,
solving for we get:
Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that
which gives
(b).
The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,
the vertical component of the velocity is
which gives a speed of