9. A ball is thrown straight up with an initial speed of 30 m/s. How long will it take to reach the top of its trajectory, and high will the ball go?

1.) Time t = 3.1 seconds

2.) Height h = 46 metres

Explanation:

given that the initial velocity U = 30 m/s

At the top of the trajectory, the final velocity V = 0

Using first equation of motion

V = U - gt

g is negative 9.81m/^2 as the object is going against the gravity.

Substitute all the parameters into the formula

0 = 30 - 9.81t

9.81t = 30

Make t the subject of formula

t = 30/9.81

t = 3.058 seconds

t = 3.1 seconds approximately

Therefore, it will take 3.1 seconds to reach to reach the top of its trajectory.

2.) The height it will go can be calculated by using second equation of motion

h = ut - 1/2gt^2

Substitutes U, g and t into the formula

h = 30 (3.1) - 1/2 * 9.8 * 3.1^2

h = 93 - 47.089

h = 45.911 m

It will go 46 metres approximately high.