A ball is thrown vertically upward with a speed of +19.0 m/s.?

Question

A ball is thrown vertically upward with a speed of +19.0 m/s.

(a) How high does it rise?

(b) How long does it take to reach its highest point?

(c) How long does the ball take to hit the ground after it reaches its highest point?

(d) What is its velocity when it returns to the level from which it started?

Answer 1

(a) V² = U² +2*a*s

0 = 19² + 2*(-9.81) *s

0 =  361- 19.62*s

19.62s = 361

s = 361/19.62

s = 18.4m

Height acheived = 18.4m

b) V = U + a*t

0 = 19 + (-9.81) t

9.81t = 19

t = 1

1.94 seconds.

c) Time up = time down

Answer = 1.94seconds

d) It will hit the ground with the same velocity with which it was thrown = 19m/s

.

Answer 2

Do a search for.  Hangtime versus Velocity.

The first link will have the formulas you can use

To determine that.

I use it for my tennisball canon.