Answer to Question #125678 in Physics for crystal

Question

Sanna rolls a ball up to another person on a smooth ramp 19.6 m above her. The
ball reaches the other person’s hands when it is travelling 4.9 m/s uphill. If the
ramp angle slows the ball down by 3.7 m/s each second it travels up the ramp, find
the initial velocity of the ball.

Answer 1

According to the kinematic formula

$h=\frac{v_f^2-v_i^2}{2a}$

We have

$2ah=v_f^2-v_i^2\to v_i^2=v_f^2-2ah$

Each second the speed is decreased by $3.7 m/s$ . Acceleration is the change of velocity with time. So,

$a=-3.7/1=-3.7m/s^2$

$v_i^2=v_f^2-2ah\to v_i=\sqrt{v_f^2-2ah}=\sqrt{4.9^2-2\cdot (-3.7)\cdot 19.6}=13m/s$