use the equation of motion: y(t)=y0+v0t-1/2gt^2 y(t)=height of projectile at any time, t y0=initial height = 34.6 m v0=initial speed (to be determined) t=time of flight we are told that the ball hits the ground (y=0) at t=4, so we have 0=34.6+v0(4s)-1/2 (9.8m/s/s)(4s)^2 0=34.6+4v0-78.4 4v0=43.8 v0=10.95m/s

Reference :-

Secure Information

Content will be erased after question is completed.

Enter the email address associated with your account, and we will email you a link to reset your password.

Forgot your password?

Sign Up