I've had alot of trouble solving this question. if anyone can explain how to solve it with steps and explanations that would be amazing!! Thank you!
An Olympic high-jumper runs toward a horizontal bar with a height of 2.00 m above
the ground. He begins his jump with a take-off speed of 10.2 m/s at an angle of 38.3°
from the ground and achieves his maximum height when he reaches the bar. By how
many centimetres does he clear the bar?
An Olympic high-jumper runs toward a horizontal bar with a height of 2.00 m above
the ground. He begins his jump with a take-off speed of 10.2 m/s at an angle of 38.3°
from the ground and achieves his maximum height when he reaches the bar. By how
many centimetres does he clear the bar?
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y-yo=(vosinΘ)t-1/2*gt²
10.2*sin38.3°=4.9t; t=1.29s
At maximum height t=1.29/2=0.645s
y=(vosinΘ)t-1/2*gt²
=10.2*sin38.3°*0.645-4.9*(0.645)²=2.03…
2.039-2.0=0.039m=3.9cm
10.2*sin38.3°=4.9t; t=1.29s
At maximum height t=1.29/2=0.645s
y=(vosinΘ)t-1/2*gt²
=10.2*sin38.3°*0.645-4.9*(0.645)²=2.03…
2.039-2.0=0.039m=3.9cm