Question

A baseball is thrown at an angle of 29 relative to the ground at a speed of 24.3 m/s. The ball is caught 51.0463 m from the thrower. The acceleration due to gravity is 9.81 m/s2. How long is it in the air? Answer in units of s.

Answer 1

Answer:

T = 2.4 s

Explanation:

given,

angle at which ball is thrown = 29°

speed relative to ground = 24.3 m/s

ball is caught at a distance = 51.0463

acceleration due to gravity = 9.8 m/s²

time for which ball was in the air = ?

now,

velocity of ball in x-direction

V_x =  v cos θ

V_x =  24.3 x cos 29°

V_x = 21.25 m/s

velocity in y direction

V_y =  v sin θ

V_y =  24.3 x sin 29°

V_y = 11.78 m/s

distance on the ground when ball will reach maximum height

x = 51.0463/2 = 25.52 m

at top most point velocity is equal to zero

time for which the ball was in air

v = u + a t

0 = 11.78 - 9.8 t

$t = \dfrac{11.78}{9.8}$

t = 1.20

this time is taken to travel half distance

total time = 2 x 1.20

T = 2.4 s

time for which ball was in air is T = 2.4 s