A ball is dropped from a window 44.1 m above the sidewalk. Air resistance is ignored. How long does it take to reach the sidewalk? Enter the time in seconds.
One throw, one whole story: up from a raised point, momentarily at rest at the top, down past the launch height, landing below. You've already seen every tool this flight needs. A stated sign choice; the time and height of the rise; the up-and-down symmetry; the dropped-object equations. Chained in order, they give the flight's total time and its landing velocity.
Stage 1 is the rise. Choose the positive direction once and state it: up positive. With v = 0 at the top, t = (v − v₀)/a gives the rise time, and Δx = (v² − v₀²)/(2a) gives the extra height above the launch point. These are the same two calculations you've already practiced.
Stage 2 is the fall. From the top, the object is a dropped object: it starts from rest at the total height, the launch height plus the extra rise. Stage 2 gets the drop convention (down positive, from the top), and you state that switch when you make it. Under the drop convention, t₂ = √(2Δx/g) gives the fall time, and v = g × t₂ gives the landing speed, directed downward.
Total time in the air = rise time + fall time. The landing velocity is the stage-2 speed: with up positive, it carries a minus sign.
A ball is thrown straight up at 14.7 m/s from a walkway 19.6 m above the ground, and lands on the ground below. Air resistance is ignored. Find the total time in the air and the landing velocity.
Write down the values in the question:
v₀ = +14.7 m/s (taking up as positive)
a = −9.8 m/s²
launch point 19.6 m above the ground
Write down the equations, one per stage:
rise: t₁ = (0 − v₀)/a and Δx = (0 − v₀²)/(2a)
fall (drop convention: down positive from the top): t₂ = √(2Δx/g), falling from rest, then v = g × t₂
Substitute in the values:
t₁ = (0 − 14.7)/(−9.8) = 1.5 s
extra height: Δx = (0 − 14.7²)/(2 × (−9.8)) = 11.025 m
top of the flight: 19.6 + 11.025 = 30.625 m above the ground, momentarily at rest
t₂ = √(2 × 30.625 / 9.8) = √6.25 = 2.5 s
Simplify:
total time: t = 1.5 + 2.5 = 4.0 s
landing speed: v = 9.8 × 2.5 = 24.5 m/s, directed downward (−24.5 m/s with up positive)
A ball is thrown straight up at 19.6 m/s from a balcony 24.5 m above the street, and lands on the street below. Air resistance is ignored. For how long is the ball in the air? Enter the time in seconds.
When the flight starts and ends at the same height, symmetry is the shortcut: the fall retraces the rise — equal time, equal speed, direction reversed.
A ball is kicked straight up at 24.5 m/s from ground level and lands back at the height it left from. Air resistance is ignored. Find the total time in the air and the landing velocity.
Write down the values in the question:
v₀ = +24.5 m/s (taking up as positive)
a = −9.8 m/s²
launch and landing at the same height
Write down the equations (symmetry is the shortcut when the heights match):
t = t₁ + t₂, with t₂ = t₁ and t₁ = (0 − v₀)/a
landing speed = launch speed, direction reversed (matched heights)
Substitute in the values:
t₁ = (0 − 24.5)/(−9.8) = 2.5 s
t₂ = 2.5 s
Simplify:
total time: t = 2.5 + 2.5 = 5.0 s
landing speed: 24.5 m/s, directed downward (−24.5 m/s with up positive)
A stone is flicked straight up at 9.8 m/s from a bridge deck 14.7 m above the water, and drops into the water below. Air resistance is ignored. What is its speed when it hits the water? Enter the speed in m/s.
A ball is tossed straight up at 4.9 m/s from the end of a pier 29.4 m above the sea, and falls into the sea. Air resistance is ignored. How long after launch does it hit the water? Enter the time in seconds.
A ball is thrown straight up at 24.5 m/s from a cliff ledge 29.4 m above the beach, and lands on the beach. Air resistance is ignored. For how long is the ball in the air? Enter the time in seconds.
A ball is thrown straight up at 29.4 m/s from a rooftop terrace 34.3 m above the street, and lands on the street. Air resistance is ignored. What is its speed just before landing? Enter the speed in m/s.
A ball is launched straight up at 39.2 m/s from ground level and lands back at the height it left from. Air resistance is ignored. For how long is it in the air? Enter the time in seconds.