6.) where is the ball traveling the fastest? why? on the way up or when its because there is no air…

6.) where is the ball traveling the fastest? why? on the way up or when its because there is no air resistance. 7.) where is the ball traveling the slowest? why? at the peak because gra pulling it down 8.) what component of velocity stays constant throughout the whole path of a projectile? the h 9.) a golf ball was driven down the fairway. a.) if the ball had less horizontal velocity, how long would it take the ball to land? it will b.) if the ball had less horizontal velocity, what happens to the range? it will change 10.) diagram what happens to velocity and acceleration as an object freefalls. constant downward 11.) what is the acceleration of a car that maintains a constant velocity of 30 m/s for 3 seconds? 12.)what is the acceleration at the highest point of a ball that is tossed into the air? 10 m/s 13.) if george drops a ball off a bridge, and jerry throws a ball off a bridge, who’s ball has a greater acceleration? all the same 0m/s² 14.) a baseball is thrown perfectly horizontal at a 20 m/s. a 2nd baseball is thrown from the same horizontally, but with twice the speed. a.) compare the times each baseball stayed in the air. same time 15.) a stone is kicked horizontally at a speed of 2.5 m/s from the top of a cliff 33 m high. a.) how long does the stone take to reach the bottom? dy = 1/2at² 33 = 1/2(-10)t² b.) how far from the cliff does the stone hit? dx = vx t
Answer
Explanation:
Step1: Analyze vertical - motion formula for time
The vertical - displacement formula for free - fall is $d_y=\frac{1}{2}at^{2}$, where $d_y$ is the vertical displacement, $a$ is the acceleration due to gravity ($a = - 9.8\ m/s^{2}\approx - 10\ m/s^{2}$), and $t$ is the time. For a stone kicked horizontally from a cliff of height $h = 33\ m$, in the vertical direction, the initial vertical velocity $v_{0y}=0\ m/s$. So, $h=\frac{1}{2}gt^{2}$, and we can solve for $t$. $33=\frac{1}{2}(10)t^{2}$ (taking magnitude of $g = 10\ m/s^{2}$), then $t^{2}=\frac{33\times2}{10}=6.6$, and $t=\sqrt{6.6}\approx 2.57\ s$.
Step2: Calculate horizontal distance
The horizontal motion of a projectile is a uniform - motion with constant horizontal velocity $v_x$. The horizontal distance formula is $d_x = v_x t$. Given $v_x = 2.5\ m/s$ and $t\approx2.57\ s$, then $d_x=2.5\times2.57 = 6.425\ m$.
Answer:
15.A. $t\approx2.57\ s$ 15.B. $d_x\approx6.43\ m$