with what i know the ball is thown at a pace of 45.0m and lands 24.0m from base so lets draw the big 5 diagrams and put them together and then we find time because its what needed to evaluate the position the ball is in.
Ok let's solve this doing it the algebra way then the calculus way. Finding TIME: Given: acceleration = -9.81 m/s-2 initial velocity = 25 m/s final velocity (at the top of the trajectory, the highest elevation) = 0 time = ? Formula: a = Δv/Δt Rearranged: Δt = Δv/a = -25/-9.81 = 2.54842 seconds (multiply by two since this is the time taken to reach the "top" of the trajectory) So t = 5.09684 s = 5.1 seconds (2 sigdigs) Finding MAXIMUM HEIGHT: Let's use d = vft - 0.5at2 NOTE: use half of the time of 5.09684 seconds since it will take half this time to reach the MAX HEIGHT d = (0) - 0.5(-9.81)(2.54842)2 = 32m ---- Calculus way: Finding TIME: a = -9.81 Antiderivative: v = -9.81t + 25 Antiderivative: d = -4.905t2 + 25t (solve for t when d = 0 (the projectile is on the ground)) t = 0 s, 5.1 s (omit 0 since this is the starting time when d = 0) t = 5.1 seconds Finding MAXIMUM HEIGHT When v = 0, the ball will be at max height. Sub this t-value into the distance formula. v = -9.81t +25 = 0 t = 2.5484 s d = -4.905(2.5484)2 + 25(2.5484) = 32 m
9.17 secs
If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t^2. If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t^2.
Objects fall to the ground because of the force of gravity.
Because of the force of gravity
A ball is thrown vertically upward with an initial speed of 20m/s. Two second later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24m/s. At what height above the release point will the ball and stone pass each other?
After just over three and a quarter seconds.
If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t - 16t^2. If a ball is thrown vertically upward with a velocity of 160 ft/s, then its height after t seconds is s = 160t − 16t^2.
Objects fall to the ground because of the force of gravity.
Because of the force of gravity
A ball is thrown vertically upward with an initial speed of 20m/s. Two second later, a stone is thrown vertically (from the same initial height as the ball) with an initial speed of 24m/s. At what height above the release point will the ball and stone pass each other?
After just over three and a quarter seconds.
A ball thrown vertically upward returns to the starting point in 8 seconds.-- Its velocity was upward for 4 seconds and downward for the other 4 seconds.-- Its velocity was zero at the turning point, exactly 4 seconds after leaving the hand.-- During the first 4 seconds, gravitational acceleration reduced the magnitude of its upward velocity by(9.8 meters/second2) x (4 seconds) = 39.2 meters per second-- So that had to be the magnitude of its initial upward velocity.
100
Vertical means straight up and down,aligned with the direction of gravity. Vertically upward means vertical in the up direction. So if you jump directly straight up that is vertically upward; when you land you are travelling vertically downward.
a cricket ball is projected vertically upward direction.what kind of acceleration is acting on it
If a ball is thrown horizontally from a window on the second floor of a building, the vertical component of its initial velocity is zero.
because if gravity
OW! Not long enough!