Answer to Question #311062 in Physics for Cee

Question

A ball is thrown vertically upwards with an initial velocity of 30 m/s.

Using a time step of 0.02 s up to 6.20 s, plot the graph of the vertical distance versus time for this ball.

Hint ; Motion under gravity is described by the equation : 𝑣𝑦 = 𝑣𝑜𝑦𝑡 +1/2𝑔𝑡^2

and gravitational acceleration 𝑔 is here taken as negative.

(i) To what maximum height does the ball rise?

(ii) What is the index of time at maximum height?

(iii) How long does it take the ball to ascend to maximum height?

(iv) How long does it take the ball to hit the ground?

(v) What happens to the ball if the sign for gravitational acceleration is taken as positive?

Answer 1

(i) To what maximum height does the ball rise?

$$h_{\max}=46\:\rm m$$

(iii) How long does it take the ball to ascend to maximum height?

$$t_1=3\:\rm s$$

(iv) How long does it take the ball to hit the ground?

$$t_2=6.1\:\rm s$$

(v) What happens to the ball if the sign for gravitational acceleration is taken as positive?

It's velocity would increase.