Question

A baseball is thrown at 40 m/s at an angle of 20° with respect to horizontal from a height of 1.5 m. How far does it travel before hitting the ground?

Answer 1

Range of projectile motion is given by:

R = V0x*T

V0x = Initial horizontal velocity = V0*cos A = 40*cos 20 deg = 37.58 m/sec

T = time period of motion = ?

to find the time period, Using 2nd kinematic equation in vertical direction

h = V0y*T + (1/2)*ay*T^2

ay = acceleration due to gravity = -9.81 m/sec^2

V0y = Initial vertical velocity = 40*sin 20 deg = 13.68 m/sec

h = Vertical distance traveled = -1.5 m

So

-1.5 = 13.68*T - 0.5*9.81*T^2

4.905*T^2 - 13.68*T - 1.5 = 0

Solving above quadratic equation:

T = [13.68 + sqrt (13.68^2 + 4*4.905*1.5)]/(2*4.905) = 2.895 sec

See that we have only used +ve root of equation, Since time cannot be negative.

So, Now range of baseball will be

R = V0x*T

R = 37.58*2.895 = 108.794 m

R = 108.8 m

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