A student standing on the ground throws a ball straight up. The ball leaves the

Question

A student standing on the ground throws a ball straight up. The ball leaves the student's hand with a speed of 19.0 m/s when the hand is 2.50 m above the ground. Part A How long is the ball in the air before it hits the ground? (The student moves her hand out of the way.) You may want to review (Pages 49 -51) For help with math skills, you may want to review: Quadratic Equations Express your answer with the appropriate units For general problem-solving tips and strategies for this topic, you may want to view a Video Tutor Solution of Time in the air for a tossed ball. alue Units Submit Previous Answers Request Answer x Incorrect: Try Again

Explanation / Answer

Answer First Part A :

For the ball to be in air before it strikes the ground,we need to calculate the time of ball's air travel.

u = initial velocity = 19 m/s

g = acceleration due to gravity = 9.8 m/s2

The time of flight = 2* u / g = 2*19 / 9.8 = 3.88 seconds

3.88 seconds is the time taken by the ball in throwing from "to and fro - position of hand of boy".

T1 = Time of flight from to and fro of boy's hand position= 3.88 seconds.

Now we need to calculate the time it takes to reach to ground from height of Boy's hand =T2  

So for calculation of T2 , we will calculate u1 (initial velocity while return at hand position) from use the formula v12 - u12 = 2 g S1 = (0)2 - u12 = 2*(-9.8) *2.5 = - u12 = - 2*9.8 *2.5 =-49 = u12=49 this means u1=7 m/s

So from v1= u1 +(- g*t) = 0= 7+(- 9.8 * T2)

-7 = - 9.8 T2   OR T2 = 7 / 9.8 = 0.71 second

Total Tome of Flight = T1 + T2 = 3.88 + 0.71 = 4.59 seconds

S =height of boy's hand

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