Question

30) A foul ball is hit straight up into the air with a speed of 30.0 m/s. (a) Calculate the time required for the ball to rise to its maximum height. (b) Calculate the maximum height reached by the ball. (c) Determine the time at which the ball pass a point 25.0 m above the point of contact between the bat and ball. (d) Explain why t

Answer 1

(c) Remember that when you square root, you get two answers. The ball will be at 25 meters two times. Think of a parabola.

Step-by-step explanation:

if you use yf = yi +vi(t)+ 1/2(a)(t^2),

you'll get -4.9t^2 +30t - 25 = 0, using the quadratic formula you will end up with -30 +/-
`\sqrt{410` all divided by -9.8

t= 1 and 5.1

6

Answer 2

a) 3.05 seconds

b) 45.87 m

c) 0.995 seconds

Step-by-step explanation:

t = Time taken

u = Initial velocity = 30 m/s

v = Final velocity = 0

s = Displacement

a = Acceleration = 9.81 m/s² (positive downward, negative upward)

a) Time taken by the ball to reach maximum height

`v=u+at\\\Rightarrow t=(v-u)/(a)\\\Rightarrow t=(0-30)/(-9.81)\\\Rightarrow t=3.05\ s`

Time taken by the ball to reach maximum height is 3.05 seconds

b) Maximum height

`v^2-u^2=2as\\\Rightarrow s=(v^2-u^2)/(2a)\\\Rightarrow s=(0^2-30^2)/(2* -9.81)\\\Rightarrow s=45.87\ m`

Maximum height reached by the ball is 45.87 m

c) If s = 25 m

`v^2-u^2=2as\\\Rightarrow v^2=2as+u^2\\\Rightarrow v^2=2* -9.81* 25+30^2\\\Rightarrow v=√(2* -9.81* 25+30^2)\\\Rightarrow v=20.24\ m/s`

Velocity of the ball at 25 m height is 20.24 m/s

`v=u+at\\\Rightarrow t=(v-u)/(a)\\\Rightarrow t=(20.24-30)/(-9.81)\\\Rightarrow t=0.995\ s`

Time taken by the ball to reach a point 25.0 m above the point of contact between the bat and ball is 0.995 seconds