For general projectile motion, when the projectile is at the highest point of it

Question

For general projectile motion, when the projectile is at the highest point of its trajectory: a. Its acceleration is zero. b. Its velocity is perpendicular to the acceleration c. Its velocity and acceleration are both zero d. The horizontal component of its velocity is zero e. The horizontal and vertical components of its velocity are zero Approximately how many pennies would you have to stack to reach an average 2 meter ceiling? a. 2 times 10^2 b. 2 times 10^3 c 2 times 10^4 d. 2 times 10^5 e. 2 times 10^6 Car A with a mass of 2 m is traveling west at a speed of v. Car B with a mass of m is traveling west at a speed of 2v. Both cars slam on their brakes at the same time, the cars to a stop. Assuming the coefficient of static between the wheels of the car and the road is the same, which are travels while stopping? a. They travel the same distance b. car A travels farther c. Car B travels farther d. Impossible to tell without knowing v e. Impossible to tell without knowing m Suppose that a car traveling to the west (the-x direction) begins to slow down as it approaches a traffic light. Which statement concerning its acceleration in the x direction is correct? a. Both its acceleration and its velocity are positive b. Both its acceleration and its velocity are negative c Its acceleration is positive but its velocity is negative d. Its acceleration is negative but its velocity is positive e. Nothing certain can be said about the relationship between its acceleration and velocity

Explanation / Answer

1)

At the highest point, its vertical velocity becomes 0 and thus only horizontal velocity remains.

So, velocity becomes perpendicular to the acceleration (which is vertically downwards)

So, options b) and d) are correct

2)

Height of 1 penny = 1 mm = 0.001 m

So, number of pennies required = 2/0.001 = 2000 = 2*10^3 <----- option b

3)

Using the equation of motion,

v^2 = u^2 + 2as

Now, for each body,

a = acceleration = u*mg/m = u*g <----- u = coefficient of friction

v = final velocity = 0

s = distance traveled

So, for the car A,

0^2 = v^2 - 2*(ug)*s1

So, s1 = v^2/2ug

Similalrly, for the car B:

0^2 = (2v)^2 - 2*(ug)*s2

So, s2 = 4v^2/2ug = 4*(v^2/2ug) = 4*s1

So, car B travels greater than car A . <------ option c

4)

As it slows down, so its acceleration must be opposite to the direction of velocity

So, acceleration is positive though the velocity is negative

So, option c) is correct

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