Question

a base ball leaves a bat with an initial speed of 37m/s at an angle of 53.1°. find position of the ball when time =2 second​

Answer 1

Answer:

Explanation:

Assuming the baseball is in projectile,

          Horizontal range R=u²*sin^theta/g

                                          =37²*sin53.1/10(Taking g as 10 m/s²)

                                           =1369*0.8/10

                                           =1095.2/10

                                           =109.52 m (Note: Value is different is g is 9.8)

Answer 2

The position of ball at 2 secs is x = 44.4m and y = 39.6m.

Given:

Initial speed at which a baseball leaves = 37m/s

Angle at which it leaves = 53.1°

To Find:

Position of the ball when time = 2 second​s

Solution:

x = (vcos53°)t

=> x = 3/5 x 37 x 2

=> x = 44.4m

By second equation of motion:

y = (vsin53°)t - g$t^{2}$/2

=> y = 37x2x4/5 - 5x$2^{2}$

=> y = 39.6m

Hence, the position of ball at 2 secs is x = 44.4m and y = 39.6m.

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