If the velocity of a body varied uniformly from 10 ms-1 to 25 ms-1 during a time interval of 5s what was the acceleration of the body?

For vertical motion, use the following kinematics equation:

H(t) = X + Vt + 0.5At²

H(t) is the height of the ball at any point in time t for t ≥ 0s

X is the initial height

V is the initial vertical velocity

A is the constant vertical acceleration

Given values:

X = 1.4m

V = 0m/s (starting from free fall)

A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)

Plug in these values to get H(t):

H(t) = 1.4 + 0t - 4.905t²

H(t) = 1.4 - 4.905t²

We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:

1.4 - 4.905t² = 0

4.905t² = 1.4

t² = 0.2854

t = ±0.5342s

Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))

t = 0.53s

Answer:0.53s

Explanation:

Answer:

9.43 m/s

Explanation:

The law of conservation of energy requires that the total energy before and after transformation must be equal. Kinetic energy of car and rocket combined equals the parachute energy and that of car.

Final energy of car will be 66120-36733=29,387 J

Kinetic energy, and making v the subject of the formula then

Where m is the mass and v is velocity.

Substituting 29387 J for KE and 661 kg for m then