Question

A pitcher throws a baseball straight up, with an initial speed of 25 m/s. (a) How long does it take to reach the highest point? (b) How high does the ball rise above its release point? (c) How long will it take for the ball to reach a point 25 m above its release point?

Answer 1

a) 2.55 seconds

b) 31.86 m

c) 1.4 seconds

Step-by-step explanation:

t = Time taken

u = Initial velocity = 25 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

a)

`v=u+at\\\Rightarrow 0=25-9.81* t\\\Rightarrow (-25)/(-9.81)=t\\\Rightarrow t=2.55 \s`

Time taken by the ball to reach the highest point is 2.55 seconds

b)

`s=ut+(1)/(2)at^2\\\Rightarrow s=25* 2.55+(1)/(2)* -9.81* 2.55^2\\\Rightarrow s=31.86\ m`

The highest point reached by the ball above its release point is 31.86 m

c)

`v^2-u^2=2as\\\Rightarrow v=√(2as+u^2)\\\Rightarrow v=√(2* -9.81* 25+25^2)\\\Rightarrow v=11.6\ m/s`

`v=u+at\\\Rightarrow 11.6=25-9.81* t\\\Rightarrow (11.6-25)/(-9.81)=t\\\Rightarrow t=1.4 \s`

It takes 1.4 seconds to reach a point of 25 m above its release point