A coin relased at the top of a tower hits the ground after falling 1.5 s. What is the speed of the coin as it hits the ground? (Disregard air resistance
vfinal=g*time and g=9.8m/s^2
thank you..but i don't understand how that will i can solve it still.... so i jus multiply
(32.0 m)(2.00s)(9.8 m/s^2) n i will have my answer?
please help...im so confused
wait im sorry.. ignore that last response...i meant i jus multiply 9.8m/s^2)(1.5 s)...n i will have my answer
To find the speed of the coin as it hits the ground, we can use the equations of motion. One equation that relates distance, time, and acceleration is:
Δd = v0 * t + (1/2) * a * t^2
Where:
Δd = change in distance (in this case, the distance the coin falls from the top of the tower to the ground)
v0 = initial velocity (the velocity of the coin when it was released)
t = time it takes for the coin to hit the ground (given as 1.5 s)
a = acceleration due to gravity (approximately -9.8 m/s², negative because it acts downwards)
In this case, we want to find the final velocity (speed) of the coin when it hits the ground. The final velocity is related to the initial velocity, acceleration, and time by the equation:
v = v0 + a * t
Since the coin was released from rest (v0 = 0), substituting this value into the equation above gives:
v = 0 + (-9.8 m/s²) * t
Now we can substitute the given value of t (1.5 s) into the equation:
v = 0 + (-9.8 m/s²) * 1.5 s
Calculating the expression:
v = -14.7 m/s
The speed of the coin as it hits the ground, disregarding air resistance, is 14.7 m/s in the downward direction. The negative sign indicates that the velocity is directed downwards.