A ball is thrown vertically upward from the ground. On the graphs below sketch t

Question

A ball is thrown vertically upward from the ground. On the graphs below sketch the acceleration versus time, the velocity versus time, and the displacement versus time A ball is thrown with an initial velocity v_0 =23.0 m/s at an angle theta= 53 0degree above the horizontal The ball is released from a rooftop that is a height y_0 = 31.0 m above ground level. Determine the time of flight for the ball to reach the ground.How far from the base of the building will the ball strike the ground A rider on cart throws a ball straight up into the air by his observations. What will an observe on the ground as the path of the ball. Describe the motion completely. Compare the vertically velocity of the ball as seen by the two observers Does the speed of the cart have any effect on the height of the ball attains or the time the ball is in the air?

Explanation / Answer

4.A

Since we know that when a ball is projected under gravity ,it moves under constant action of gravitational pull of earth (=mg; vertically downward). There fore acceleration of particle will be constant (= 9.8 m/s^2; vertically downward direction)

.....So graph of a-t curve will be || to time axis and it will intersect 'a' axis at -9.8 (I have chosen vertical downward direction negative)

.....formula for instantaneous velocity (upward direction positive) will be

V = u- gt

So V - T curve will be a St.line with negative slope and it will intersect 'time' axis at T= u/g

Formula for vertical displacement will be

Y= UT - 1/2 gt2 ( equation of a parabola , which will pass through origine and will be convex upward . it will intersect time axis at two points [ first at t=0 and second at t= u/g

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