A STUDENT THROWS A BALL AT THE SPEED 8 M/S AT THE ANGLE OF 40DEGREES ABOVE THE H

Question

A STUDENT THROWS A BALL AT THE SPEED 8 M/S AT THE ANGLE OF 40DEGREES ABOVE THE HORIZONTAL AND SECOND STUDENT CATCHES THE BALL1.2 SECONDS LATER. THE BALL IS THROWN FROM AN INITIAL HEIGHT 1.5 MABOVE THE GROUND. A. AT WHAT HEIGHT IS THE BALL CAUGHT B. IS THE BALL CAUGHT BEFORE OR AFTER IT REACHES MAX HEIGHT?SHOW WORK. C. FIND THE VELOCITY (MAGNITUDE AND DIRECTION) OF THE BALL ASIT IS CAUGHT. SKETCH THE FINAL VELOCITY AND CLEARLY LABEL ITSMAGNITUDE AND ANGLE. A STUDENT THROWS A BALL AT THE SPEED 8 M/S AT THE ANGLE OF 40DEGREES ABOVE THE HORIZONTAL AND SECOND STUDENT CATCHES THE BALL1.2 SECONDS LATER. THE BALL IS THROWN FROM AN INITIAL HEIGHT 1.5 MABOVE THE GROUND. A. AT WHAT HEIGHT IS THE BALL CAUGHT B. IS THE BALL CAUGHT BEFORE OR AFTER IT REACHES MAX HEIGHT?SHOW WORK. C. FIND THE VELOCITY (MAGNITUDE AND DIRECTION) OF THE BALL ASIT IS CAUGHT. SKETCH THE FINAL VELOCITY AND CLEARLY LABEL ITSMAGNITUDE AND ANGLE.

Explanation / Answer

initial vertical velocity=8*sin40 let vertical displacement=h vertical acceleration=g=9.8 t=1.2 h=-u*t+(1/2)*g*t2h=-8*sin40*1.2+(1/2)*9.8*1.22 h=-6.17+7.05 h=0.88 the ball is caught (1.5-0.88)=0.62 m above the ground time taken to reach max height=T v=u-g*T 0=8*sin40-9.8*T T=0.52 the ball is caught after it reaches it's max height horizontal velocity remains constant since there is no horizontalacceleration=8*cos40=6.13 m/s vertical velocity after the ball fell for (1.2-0.52)=0.62s from themax height=v v=u+g*t v=0+9.8*0.62 v=6.66 velocity when the ball iscaught=(6.132+6.662)=9.05 m/s

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