20 pints) ming no air drag ic uakes 2.10 seconds to reach masimam heigh At half

Question

20 pints) ming no air drag ic uakes 2.10 seconds to reach masimam heigh At half of that time, 1,0S seconds (for A softhall piächer theows a hall straight upward as hand as she can. Acconding to your fopwarch. it takes exactly 4.20 scconds for the ball to fall back to the pitcher's hand Insiructon s use oniy) a) What was the speed at which the pitcher theew the ball upward? (b) What is the maximum height reached by the hall? how high is the ball (ls it half-way to maxinum heighe) (20 pomh) Aprojectileis launched with an initial velocity of(10 m/14(12m/ej ak shown below, Whst is the velocity of the projectile when it highest point 7 Explain and draw its veloeity at its highest point. ( pt) 44 reaches its (20 pels. Obje" B starts from rest as the onpa sad accelnates at a (es (+ a hal (s) of B at say time t (b) the speed of B ar azy time t; (e) the position of B at aay time t 5 Pee) , ave an example of s cas, were the velocity of 8a object is ia the emente directie of scoelerscios,OR sucha thing is inmpossibie, explais why. the

Explanation / Answer

We will do 4 parts as per Chegg's guidelines:

Q.3 a) time to reach max height = 2.10 seconds,

u (initial velocity) = 9.8 (2.10) = 20.58 m/s

b) max height = 1/2 (9.8) ( 2.10)^2=21.609 m

c)h = 1/2 (9.8) ( 1.05)^2= 5.4 m apprx

Q. 4 At the max height, the vertical component of velocity reduces to 0 ,

V = sqroot ( vx^2+ vy^2) = sqroot ( 10^2 + 0^2) = 10 m/s

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